PJ's Chess TOUNAMENT

First of all , we should understand that

(number of matches played) = (number of points distributed)

Let the total number of players be n

Total matches played = $1+2+3+.....+(n-2) +(n-1)$ = $\frac{n * (n-1)}{2}$

Thus $\frac{n*(n-1)}{2} =$ $\frac{35}{2} +(n-4) x$ ; where x is the score of each of the remaining player!

$\frac{n^2 - n - 35}{2 (n-4)} = x$

Now x is either a multiple of $0.5$ or an integer. But we can completely assure that $2x$ is an integer.

$2x\ =$ $\frac{n^2 - n - 35 }{n-4}$ $\Rightarrow INTEGER$

Dividing by long division method , we get $2x =$ $(n+3) -$ $\frac{23}{n-4}$

Now $(n+3)$ is already an integer . So for $2x$ to be an integer $\frac {23}{n-4}$ should be an integer.

That is possible only if $(n-4) =$ 1 or $(n-4)=23$

If $(n-4)$ is 1 then x comes out to be -ve , which is not possible.

Thus $(n-4)\ = 23$ And so $n =$ $\boxed {27}$