Reciprocal sum is exactly 1 now

First, we can prove that $a \leq 4$ .

Because, $a \leq b \leq c \leq d$ we deduce that $\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c} \geq \frac{1}{d}$ which implies that $1=\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \leq \frac{4}{a}$ , so $a \leq 4$ . We also have that $1=\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \geq \frac{4}{d}$ , so $d \geq 4$

Now, - for $a=4$ , the equation becomes . $\frac{3}{4}=\frac{1}{b} + \frac{1}{c} + \frac{1}{d}$ , with the same method we have that $\frac{3}{b} \ge \frac{3}{4} \implies b \le 4 \implies b=4$ because $b \geq a = 4$ . Then c=4 and d=4 with the same method. $\implies S_1=\boxed{\boxed{1}}$ solution.

for $a=3$ the equation becomes . $\frac{2}{3}=\frac{1}{b} + \frac{1}{c} + \frac{1}{d}$ , applying the same method, we get b=4 or b=3. With b=3 the equation becomes $\frac{1}{3}=\frac{1}{c} + \frac{1}{d}$ , and because $\frac{1}{c} + \frac{1}{d} \leq \frac{2}{c}$ , $c \leq 6$ and we don't forget that $c \geq 3$ , so by testing we conclude that for c=6, d=6 and for c=4, d=12 $\implies \boxed{2}$ solutions. In addition to this, for b=4 the equation becomes $\frac{5}{12}=\frac{1}{c} + \frac{1}{d}$ so $c \leq \frac{24}{5} \implies c=4 \implies d=6 \implies \boxed{1}$ solution. $\implies S_2=\boxed{\boxed{3}}$ solutions.

for $a=2$ , the equation becomes $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{2}$ . b can't be 2 so $b \geq 3$ and $\frac{1}{2} \leq \frac{3}{b} \implies 3 \leq b \leq 6$ . After testing the long possibilities, we find : for b=6 there is $\boxed{1}$ solution for b=5 there is $\boxed{1}$ solution

for b=4 there are $\boxed{3}$ solutions

for b=3 there are $\boxed{5}$ solutions

$\implies S_3=\boxed{\boxed{10}}$ solutions.

$\implies S=\boxed{\boxed{S_1+S_2+S_3=10+3+1=14}}$ solutions.

Observe since there are only four variables, each fraction must be relatively large, and hence all of $a,b,c,d$ must be relatively small, indicative of a small number of solutions.

Since $\frac {1}{a} \geq \frac {1}{b} \geq \frac {1}{c} \geq \frac {1}{d}$ , it follows that $\frac {4}{a} \geq 1 > \frac {1}{a}$ . Thus, it is evident that $2\leq a\leq 4$ .

Case 1: Consider $a=2$ . Then since $b\leq c\leq d$ , we similarly have $\frac {3}{b} \geq \frac {1}{2}$ and $1 - \frac {1}{2} > \frac {1}{b}$ so $3\leq b\leq 6$ .

Consider $b=3$ , $\frac{1}{c}+\frac{1}{d}=\frac{1}{6}$ . Similarly to previously, $7\leq c\leq12$ . To find pairs of $(c,d)$ , we may rearrange the equation as $6c=(c-6)d$ . As is evident, the possible values for $(c,d)$ are $(c,d)=(7,42),(8,24),(9,18),(10,15),(12,12)$ , making 5 solutions.

Consider $b=4$ , $\frac{1}{c}+\frac{1}{d}=\frac{1}{4}$ ; $5\leq c\leq8$ . As previously, $(c,d)=(5,20),(6,12),(8,8)$ , making another 3 solutions.

Consider $b=5$ , $\frac{1}{c}+\frac{1}{d}=\frac{3}{10}$ ; $5\leq c\leq\frac{20}{3}$ . Again, $(c,d)=(5,10)$ , just 1 solution.

Consider $b=6$ , $\frac{1}{c}+\frac{1}{d}=\frac{1}{3}$ . Since half of $\frac{1}{3}$ is $\frac{1}{6}$ and $b=6$ the only solution is $(c,d)=(6,6)$ .

Making a total of 10 solutions for $a=2$ .

Case 2: Now consider $a=3$ . Thus $3\leq b\leq\frac{9}{2}$ .

$b=3$ , so $\frac{1}{c}+\frac{1}{d}=\frac{1}{3}$ . $(c,d) = (6,6)$ and $(4,12)$ are solutions. Hence there are 2 solutions for this branch.

$b=4$ , and $\frac{1}{c}+\frac{1}{d}=\frac{5}{12}$ . Since $\frac{24}{5}<5$ either $c=4$ or no solution. If $c=4$ , $d=6$ . 1 solution.

Case 3: Making a total of 3 solutions for $a=3$ .

Finally, when $a=4$ , all of $a,b,c,d$ must be equal to 4. This is the only solution.

This makes a total of 14 solutions.

[Slight edits for clarity - Calvin]

Before starting it is apparent that no solution can consist of all odd denominators or just one even denominator.
Beginning with the values for a :
- If a is greater than or equal to 5 or equal to 1, then a solution cannot be achieved. So a =2, 3, or 4

• If a = 4
  4 ≤ b ≤ c ≤ d
  There is 1 solution in this case. The only solution for this case is ( a , b , c , d ) = (4, 4, 4, 4)
  
  

• If a = 3
  3 ≤ b ≤ c ≤ d and b cannot be ≥ 5 in order to achieve a solution because
  1/5+1/5+1/5=0.6 <0.66
  So b = 3 or 4
  

• If a and b = 3 then 3 ≤ c ≤ d and c cannot be ≥ 7 in order to achieve a solution because
  1/7+1/7=.285… <.33
  So c =4or 6 and d =12 and 6 respectively. If c= 5 then it makes only one even possible so it is not a solution.
  There are 2 solutions. The only solutions in this case are ( a , b , c , d )=(3,3,4,12) and (3,3,6,6)

• If a =3 and b =4 then
  4 ≤ c ≤ d and c cannot be ≥ 5 in order to achieve a solution because
  1/5+1/5=.4<.41666…
  So c =4 and d =6
  There is 1 solution. The only solution in this case is ( a , b , c , d )=(3,4,4,6)

• If a =2
  2 ≤ b ≤ c ≤ d and b cannot be ≥ 7 or equal to 2 in order to achieve a solution because 1/7+1/7+1/7=.42857<.5 So b= 3, 4, 5, or 6

• If a=2 and b=3 3 ≤ c ≤ d and c cannot be ≥ 13 or less than 7 in order to achieve a solution because 1/13+1/13=.153<.16
  So c =7,8,9,10, or 12 and d=42,24,18,15,12. c = 11 does not give you an integer for d .
  There are 5 solutions. The only solutions in this case are ( a , b , c , d )=(2,3,7,42),(2,3,8,24),(2,3,9,24),(2,3,10,18),(2,3,11,15)

• If a=2 and b=4
  4 ≤ c ≤ d and c cannot be ≥ 9 or less than 5 in order to achieve a solution because 1/9+1/9=.22<.25
  So c =5,6,8, and d =20,12,8. c =7 or 9 does not give you an integer for d
  There are 3 solutions. The only solutions in this case are ( a , b , c , d )=(2,4,5,20),(2,4,6,12),(2,4,8,8)

• If a=2 and b=5
  5 ≤ c ≤ d and c cannot be ≥ 7 in order to achieve a solution because
  1/7+1/7=.285<.3
  So c =5 and d =10. c =4 or 6 does not give you an integer for d
  There is 1 solution in this case. The only solution in this case is ( a , b , c , d )=(2,5,5,10)

• If a =2 and b =6
  6 ≤ c ≤ d and c cannot be ≥7
  1/7+1/7=.285<.33
  So c =6 and d =6.
  There is 1 solution in this case. The only solution in this case is ( a , b , c , d )=(2,6,6,6)

$\Rightarrow$ Therefore, there are: * 1+2+1+5+3+1+1=14 solutions *

First of all, $a=1$ does not work: the sum on the left will be too big.

Case 1. If $a=2$ , then $b\geq 2$ . The case $b=2$ does not work, so the smallest $b$ could be is $3$ .

Case 1.1 If $a=2$ and $b=3$ , then $c\geq 3$ and $\frac{1}{2}+\frac{1}{3} +\frac{2}{c} \geq 1$ , which implies that $c\leq 12$ . One can check that $c$ could be $7,\ 8,\ 9,\ 10,$ or $12$ . So we get $5$ solutions.

Case 1.2 If $a=2$ and $b=4$ , then $c\geq 4$ and $\frac{1}{2}+\frac{1}{4} +\frac{2}{c} \geq 1$ , which implies that $c\leq 8$ . One can check that $c$ could be $5,\ 6,$ or $8$ . So we get $3$ solutions.

Case 1.3 If $a=2$ and $b=5$ , then $c\geq 5$ and $\frac{1}{2}+\frac{1}{5} +\frac{2}{c} \geq 1$ , which implies that $c\leq 6$ . One can check that $c=5$ works, so we get $1$ solution.

Case 1.4 If $a=2$ and $b=6$ , then $c=d=6$ , and we get one more solution.

Case 1.5 Note that if $a=2$ , then $b$ cannot be $7$ or larger, because we require that $\frac{1}{2}+\frac{3}{b}\geq 1.$

Case 2.1 If $a=3$ , then $b$ could be $3$ , which yields two solutions ( $c=4$ and $c=6$ )

Case 2.2 If $a=3$ and $b=4$ , this yields one solution ( $c=4$ ).

Case 2.3 If $a=3$ , then $b$ cannot be $5$ or larger, because we require that $\frac {1}{a} + \frac {3}{b} \geq 1$

Case 3. The biggest $a$ could be is $4$ , when $a=b=c=d=4$ , so one more solution.

So overall, we get $14$ solutions.