Be Logical

Beautiful problem, simple and logical,,

Its funny how the simple condition that all involved numbers are natural numbers gives us just enough equations to solve the problem, like the constraint equations in physics,

let there be 'n' positive consecutive integers on board starting from 1 initially,

then initial sum of the numbers is

$\frac { n(n+1) }{ 2 }$

now if k was removed, then sum of numbers becomes

$\frac { n(n+1) }{ 2 } -k$

and the corresponding AM is

$\frac { n(n+1) }{ 2(n-1) } -\frac { k }{ n-1 }$ which is equal to

$35+\frac { 7 }{ 17 } =\frac { (7)(86) }{ 17 }$

now isolating k in the equations we get

$\frac { n(n+1) }{ 2 } -\frac { (7)(86) }{ 17 } (n-1)=k$

Now k should be an integer, and we know n(n+1)/2 is obviously an integer , so we need

$\frac { (7)(86) }{ 17 } (n-1)\\$ to be an integer,

which is possible only if n-1 is divisible by 17, so let n-1=17c for some natural number 'c'

also we need k to be less than n, using both the constraints we see that c=4 or n=69 from which we get k =7

However i personally recommend avoid the second constraint as it is cumbersome,, as clearly we need the smallest possible solution for 1st constraint which is easily verifiable to be at c = 4

It was really fun solving this... Beautiful problem.

There is a property that i just observed and proved myself. $It\ says\ if\ AM\ of\ first\ n\ natural\ no.'s\ =\ K$

$Then, \ AM\ after\ erasing\ any\ no.\ out\ of\ it,\ say\ W \ \in [(k-\frac{1}{2}) ,(k+\frac{1}{2})]$ $\boxed{Try\ to\ prove\ this\ yourself.. It's\ easy..}$

$\Rightarrow The \ AM \ of\ no.'s\ before\ erasing\ can\ only\ be\ 35\ or\ 35.5.$

$\Rightarrow n=69\ or\ 70$

Now let the required no. be X.

$\Rightarrow X= \frac{n*(n+1)} {2} - \frac{602}{17} (n-1)\$

Since (n-1) must be a multiple of 17 gives $n=69.$

Putting it back in the equation , we get

$\boxed {X=7}$