, 2020

Gabrielle can always win using the following strategy: during each of her turns, she picks the largest two numbers on the sheet as $a$ and $b$. Using induction on $k$, we will prove that she is always allowed to do so, and that after her $k$-th turn, the sheet contains the number $2k + 1$ and $998 - 2k$ ones.

In his first turn, Ward can only pick $a = b = 1$, after which the sheet contains the number $2$ and $997$ ones. Gabrielle then picks the two largest numbers, $a = 2$ and $b = 1$, after which the sheet contains $3$ and $996$ ones. This finishes the basis $k = 1$ of the induction.

Now suppose that for some $m \ge 1$ after Gabrielle's $m$-th turn the sheet contains the number $2m + 1$ and $998 - 2m$ ones. If $998 - 2m = 0$, then Ward cannot make a move. If not, then Ward can do one of two things, either pick $a = b = 1$ or pick $a = 2m + 1$ and $b = 1$. We consider these two cases separately:

* If Ward picks $a = b = 1$, then the sheet contains the number $2m + 1$, the number $2$, and $996 - 2m$ ones. Gabrielle then picks the two largest numbers, so $a = 2m + 1$ and $b = 2$ (which is allowed since their gcd is $1$). After her turn the sheet contains the numbers $2m+3 = 2(m+1)+1$ and $996 - 2m = 998 - 2(m+1)$ ones.

* If Ward picks $a = 2m + 1$ and $b = 1$, then the sheet contains the number $2m+2$ and $997 - 2m$ ones. Gabrielle then picks the two largest numbers, so $a = 2m + 2$ and $b = 1$ (which is allowed since their gcd is $1$). Note that there is a one left, as $997 - 2m$ is odd, so not equal to $0$. After her turn the sheet contains the numbers $2m + 3 = 2(m + 1) + 1$ and $996 - 2m = 998 - 2(m + 1)$ ones.

This completes the induction.

Therefore Gabrielle can always make a move. After Gabrielle's turn $499$ the only number left on the sheet is $999$, so Ward can no longer make a move, and Gabrielle wins.