0.
We first claim that if $n$ is odd, then $n^{2k} \equiv 1 \pmod{2^{k+2}}$. This can be verified inductively: first prove it for $k=1$, and the inductive step reduces to checking that $(1+2^{k+2})^2-1$ is divisible by $2^{k+3}$.
It follows that $n^{24} = (n^3)^8$ is congruent to either 0 or 1 (mod 32), depending on whether $n$ is even or odd. Thus the desired left-hand side sum can only have remainder between 0 and 24 inclusive modulo 32, but not $-1$.
Now, 2000 is a multiple of 16 but not a multiple of 32, so $2015 \equiv -1 \pmod{32}$. Thus $2015^{2015} \equiv -1 \pmod{32}$ as well, and there are no solutions of the desired equation.