Notice $f(m,n)=(-1)^{m+n}$ Let $\pi_j$ satisfy $x_{\pi_j}=j$. Since the sum is symmetric, the sum is equal to $$\sum_{1\le i<j\le 2565} \pi(i)\pi(j) (-1)^{i+j}$$ $$=\frac 12 \sum_{i\ne j} \pi(i)\pi(j) (-1)^{i+j}$$ $$=\frac 12 \left( \left( \sum_{i=1}^{2565} (-1)^i\pi(i) \right)^2 - \sum_{i=1}^{2565} \pi(i)^2 \right)$$ Note $\left( \sum_{i=1}^{2565} (-1)^i\pi(i) \right)$ is odd, so the sum is at most $\frac 12 (1-\sum\limits_{i=1}^{2565} i^2)$ To see equality can be achieved, have $\pi^{-1}(j)=x_j$ be an odd number if and only if $j\equiv 1$ or $j\equiv 2(\bmod\; 4)$.