Add 2 or Subtract 10

Every two turns the number of stones in the pile will be divisible by four. Therefore, the first player will never be able to remove the last stone, since he will always be presented with a number that is divisible by 4.

The strategy for the second player is to keep removing 10 stones until the number of stones left is less than 10.

Then, when the number gets to 10, remove the final 10!

Let person 1 be the one who plays first, and person 2 be the one who plays second. Let $A$ be the number of "add 2 stones" moves by persons 1 and 2 combined, and $S$ be the number of "subtract 10 stones" moves by persons 1 and 2 combined.

The event "no stone is left" satisfies the equation $1000+2A-10S=0,$ or rewriting, $A=5S-500.$ Clearly, $A \geqslant 0$ , and thus $S\geqslant 100$ . Now, if $S$ is even, then $S=2k,k\in\mathbb{Z^+}$ , and $A=5(2k)-500=10k-500=10(k-50)\equiv 0 \pmod{2}$ so that $A$ is also even. Now, if $S$ is odd, so that $S=2k+1, k\in\mathbb{Z^+}$ , then $A=5(2k+1)-500=10k-495 = 5(2k-99) \equiv 1 \pmod{2}$ so that $A$ is also odd.

In both cases, we have shown that $A+S$ is always even. But $A+S$ is the total number of moves done by persons 1 and 2 combined. Therefore, this implies that if person 1 moves first, then the game ends with person 2 as the last move (the winning move). Thus the one who goes second has the winning strategy.

PS: Actually, it follows from above that even if person 2 doesn't have any strategy, person 2 will always win no matter what moves they do! (provided, of course, that person 1 moves first)