Claim: Among the $10$ proper fractions chosen, $\frac1{5}$ cannot be one of them.

Proof: Suppose not. That is suppose $\frac1{5}$ is one of the $10$ proper fractions, then from the $5^\text{st}$ bullet point, the product of half of $10$ proper fractions is equal to $\frac15 ,$ and thus the product of the other half of the $10$ proper fractions must be equal to $\frac1{10} \div \frac15 = \frac12 .$

If $\frac15$ is one of the $5$ proper fractions whose product is $\frac15,$ then the remaining $4$ proper fractions have a product of $\frac15 \div \frac15 = 1 .$ This is impossible because the product of positive proper fractions cannot be greater or equal to $1.$

If $\frac15$ is one of the $5$ proper fractions whose product is $\frac12,$ then the remaining $4$ proper fractions have a product of $\frac12 \div \frac15 = \frac52 \geqslant 1 .$ This is impossible again for the same reason.

Either way, $\frac15$ cannot be one of the $10$ proper fractions.

Claim: Among the $10$ proper fractions chosen, $\frac16, \frac17, \frac18,\frac19, \frac1{10}$ cannot be one of them as well.

Proof: It is the same argument as the proof above.

Next, let's figure out the number of such proper fractions whose denominators are at most $10.$ A simple script tells us that there are $25$ such proper fractions:

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from math import gcd
from fractions import Fraction

all_such_lowest_fractions = [ Fraction(a,b) for a in range(1,9 + 1) for b in range(a+1, 10 + 1) if gcd(a,b) == 1 ]

# Because we have established that neither 1/5, 1/6, ... , 1/10 can be one of them.
for d in range(5, 10 + 1):
    all_such_lowest_fractions .remove( Fraction(1,d) ) 

print(all_such_lowest_fractions)
print(len(all_such_lowest_fractions))
'''Output:
[Fraction(1, 2), Fraction(1, 3), Fraction(1, 4), Fraction(2, 3), Fraction(2, 5), Fraction(2, 7), Fraction(2, 9), Fraction(3, 4), Fraction(3, 5), Fraction(3, 7), Fraction(3, 8), Fraction(3, 10), Fraction(4, 5), Fraction(4, 7), Fraction(4, 9), Fraction(5, 6), Fraction(5, 7), Fraction(5, 8), Fraction(5, 9), Fraction(6, 7), Fraction(7, 8), Fraction(7, 9), Fraction(7, 10), Fraction(8, 9), Fraction(9, 10)]
25
'''

$\phantom0$

There is a total of ${25 \choose 10} = \numcomma{3268760}$ ways to select $10$ of these fractions. To add to that, for each of these $10$ -tuplets, we have to verify that they satisfy all the bullet points. That's quite a daunting task!

In order to reduce the number of cases to check, we will attempt to narrow down which of these $25$ fractions must be chosen.

Claim: $6$ of the $10$ numbers chosen must be $\dfrac45, \dfrac56, \dfrac67, \dfrac78, \dfrac89, \dfrac9{10} .$

Proof: From the $1^\text{st}$ and $7^\text{th}$ bullet point, $6$ of these proper fractions must have a product of $\dfrac1{10} \div \dfrac14 = \dfrac25 .$ Among the $25$ fractions to choose from, only the largest $6$ fractions yield such a product, $\dfrac45 \times \dfrac56 \times \dfrac67 \times \dfrac78 \times \dfrac89 \times \dfrac9{10} = \dfrac25 .$ In other words, among these $25$ fractions to choose from, $6$ of them must be $\left \{ \frac45, \frac56, \frac67, \frac78, \frac89, \frac9{10} \right \} .$

Confirmation:

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all_such_lowest_fractions.sort()
these_6_largest_fractions = all_such_lowest_fractions[-6:]
print(prod(these_6_largest_fractions ))
print(*these_6_largest_fractions )
# Output:
# 2/5 
# 4/5 5/6 6/7 7/8 8/9 9/10

$\phantom0$

Claim: The other $4$ fractions must be $\left( \frac23, \frac34, \frac57, \frac7{10} \right)$ or $\left( \frac34 , \frac35 , \frac57, \frac79 \right) .$

Proof: From the previous claim, we have determined $6$ fractions whose product is $\frac25 ,$ thus we just have to exhaustively check for all ${25-6 \choose 10-6} = {19 \choose 4} = \numcomma{3876}$ combinations of fractions that yields a product of $\frac14 .$ A simple script can instantaneously produce the answer.

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# Remove these 6 fractions from the selection first:
for d in range(5, 10 + 1):
    all_such_lowest_fractions.remove(1 - Fraction(1,d))

for this_set in combinations(all_such_lowest_fractions, 4):
    if prod(this_set)  == Fraction(1,4):
        print(*this_set)
'''Output:
2/3 3/4 5/7 7/10
3/4 3/5 5/7 7/9
'''

$\phantom0$

Thus, the two solutions are:

$\left(\dfrac23\hspace{0.5em}, \hspace{0.5em}\dfrac34\hspace{0.5em}, \hspace{0.5em}\dfrac45\hspace{0.5em}, \hspace{0.5em}\dfrac56\hspace{0.5em}, \hspace{0.5em}\dfrac57\hspace{0.5em}, \hspace{0.5em}\dfrac67\hspace{0.5em}, \hspace{0.5em}\dfrac78\hspace{0.5em}, \hspace{0.5em}\dfrac89\hspace{0.5em}, \hspace{0.5em}\dfrac7{10}\hspace{0.5em}, \hspace{0.5em}\dfrac{9}{10}\right)$ and $\left( \dfrac34 \hspace{0.5em}, \hspace{0.5em} \dfrac35\hspace{0.5em}, \hspace{0.5em} \dfrac45\hspace{0.5em}, \hspace{0.5em} \dfrac56 \hspace{0.5em}, \hspace{0.5em} \dfrac57\hspace{0.5em}, \hspace{0.5em} \dfrac67\hspace{0.5em}, \hspace{0.5em} \dfrac78\hspace{0.5em}, \hspace{0.5em} \dfrac79 \hspace{0.5em}, \hspace{0.5em} \dfrac89\hspace{0.5em}, \hspace{0.5em} \dfrac9{10}\right)$

Let's verify that these $2$ solutions satisfy all the given bullet points.

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product_of_how_many = lambda give_me_10_fractions, pick_how_many : \
                    any( prod(these_n_tuplets) == Fraction(1,pick_how_many) 
                            for these_n_tuplets in combinations(give_me_10_fractions, pick_how_many) )

are_all_the_bullet_points_true = lambda give_me_ten_fractions : \
                                all( product_of_how_many(give_me_ten_fractions, k) for k in range(2, 10 + 1) )

first_solution = [Fraction(2, 3), Fraction(3, 4), Fraction(4, 5), Fraction(5, 7), Fraction(7, 10), 
                    Fraction(5, 6), Fraction(6, 7), Fraction(7, 8), Fraction(8, 9), Fraction(9, 10)]

second_solution = [Fraction(3, 4), Fraction(3, 5), Fraction(4, 5), Fraction(5, 7), Fraction(7, 9), 
                    Fraction(5, 6), Fraction(6, 7), Fraction(7, 8), Fraction(8, 9), Fraction(9, 10)]

print( are_all_the_bullet_points_true(first_solution) )
print( are_all_the_bullet_points_true(second_solution) )
# Output:
# True
# True

$\phantom0$

The answer follows.

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final_fraction = sum(first_solution) + sum(second_solution)

print(final_fraction.numerator + final_fraction.denominator)
# 21397

$\phantom0$

What's incredible is that the results produced from the scripts are all instantaneous!

Say the fractions are, in some order, $x_1,x_2,\ldots,x_{10}$ . Obviously there are lots of ways we could pick, say, the five fractions whose product has to be one fifth ( $252$ ways, to be exact), so in most cases we can't immediately assign the fractions to their given products in most cases; but let's be more optimistic. We can say that $\prod_{i=1}^{10} x_i=\frac{1}{10}$

and, since we can choose the labelling of the $x_i$ , we can have $\prod_{i=1}^{9} x_i=\frac{1}{9}$

Comparing these, we get $x_{10}=\frac{9}{10}$ . In fact, if we assume that the fractions not included in the next product are $x_9,x_{10}$ , we'd get $\prod_{i=1}^{8} x_i=\frac{1}{8}$

and $x_9=\frac89$ . It's clear that we can keep going like this, following the pattern $x_i=\frac{i-1}{i}$ for $i>1$ .

Of course, now we see the problem: if $x_2=\frac12$ , we would need $x_1=1$ , which - although it would satisfy all the required products - is not allowed in the conditions.

Remaining optimistic (if a bit less so), let's instead say that $x_i=\frac{i-1}{i}$ for $i>2$ , and seek a pair of fractions $x_1,x_2$ not already in this list such that $x_1 x_2=\frac12$ .

Say that $x_1=\frac{p}{q}$ and $x_2=\frac{r}{s}$ , where $p,q,r,s$ are positive integers and the fractions are in lowest terms. Then we need $2pr=qs,\;\;\;1<q\le s \le 10$ (where again we're free to choose $q\le s$ .) We also need $p<q-1$ and $r<s-1$ , so as not to clash with one of the $x_i$ we've already chosen.

It's a bit of casework but not too hard to show that the unique pair of fractions satisfying these conditions are $x_1=\frac57$ and $x_2=\frac{7}{10}$ . The complete list of fractions is then $\frac57,\frac{7}{10},\frac23,\frac34,\frac45,\frac56,\frac67,\frac78,\frac89,\frac{9}{10}$ and it's easy to verify that the product of the first $n$ fractions in this list is $\frac{1}{n}$ .

The sum of these fractions is $\boxed{\frac{20123}{2520}}$ .

Note that this approach finds a set of fractions that works, but doesn't prove uniqueness (we made an assumption in calculating $x_9$ but it doesn't rule out other possibilities).