Goodbye 2015

$\frac{2015^{2}-1}{1+2+3+...+63}$

Recall the difference of two squares and the formula for the sum of an arithmetic sequence, $\frac{1}{2}n(n+1)$ .

= $\frac{(2015+1)(2015-1)}{\frac{1}{2}\times 63\times 64}$

= $\frac{2016\times 2014}{2016}$

=2014

So the solution is $\boxed{2014}$ .