Don't write all the possibilities, part 1

We have $y^2-y-x=0$ , so $y=\frac{1}{2}(1+\sqrt{1+4x})$ . Then $y$ will be a natural number exactly when $1+4x$ is an odd square (and note all odd squares are of this form) and $1\le y\le 2015$ when $5\le1+4x\le 4029^2$ (note $1+4x\ge 5$ , since $x$ is a natural number).

There are $\frac{4029-1}{2}$ odd squares between $5$ and $4029^2$ inclusive, and $\frac{4029^2-1}{4}=\frac{(4029-1)(4029+1)}{4}$ numbers of the form $4x+1$ .

Therefore, the probability that $y$ is a natural number (given $1\le y\le 2015$ ) is $\frac{4028}{2}\frac{4}{4028\cdot 4030}=\frac{1}{2015}$ , so the answer is $\boxed{2016}$ .

Rewriting the equation as y^2 = y+ x ............... or $y(y-1) = x$ we can see that as $y$ varies in all real numbers $x$ takes certain real values . Now for $x$ to be a natural number it is clear that $y$ must also be a natural number except 1 . So $y$ $\in$ ${2,3,4,.........2015}$ , for which x takes certain natural number values. So thinking in the reverse sense it is easy to understand as x runs from values $1$ to $2015*2014$ , $y$ will only take natural number values for only $2014$ values of $x$ which correspond to $y$ $\in$ ${2,3,4,.........2015}$ . So as x runs through all natural numbers from 1 to $2015*2014$ there are only $2014$ such values of x for which y is a natural number. So probability is $\frac{2014}{2015*2014}$ = $\frac{1}{2015}$ . So our answer is $1+2015$ = $2016$