The answer: $n = 12$. We numerate the piles according to their weights in increasing order. Let $S_i$ and $S'_i$ be the weights of pile number $i$ before and after removing of two stones. Then $S_1 < S_2 < \dots < S_n$ and $S'_1 > S'_2 > \dots > S'_n$. Let $S_1 - S'_1 = x$. Then $S_2 - S'_2 \ge x+2, \dots, S_n - S'_n \ge x+2(n-1)$. Now note that the heaviest stone removed from pile number one weights at least $S_1/2016$ (it is not lighter than the average weight of all remaining stones). Therefore,
$$S_1 - S'_1 = x \ge \frac{S_1}{2016} + 1.$$
Similarly, the lightest stone removed from pile number $n$ is at most $S'_n/2016$. Therefore,
$$\frac{S'_n}{2016} + 25 \ge S_n - S'_n \ge x + 2(n-1).$$
These two inequalities imply that
$$\frac{S'_n - S'_1}{2016} + 24 \ge 2(n-1) \text{ and } n < 13.$$
Let us show that $n = 12$ is possible. Below $m$ denotes the stone weighted $m$. It can be readily verified that if for $k = 1, 2, \dots, 12$ the pile number $k$ consists of $1$ stone $2k$, $2004+k$ stones $24$ and $13-k$ stones $25$ then the conditions are satisfied. Done.