Mathematical Olympiad Rioplatense

The first player $A$ has a winning strategy if $N$ is congruent to $0$ or $1$ modulo $4$, otherwise the second player $B$ has one.

Let $N = 4k + r$ where $k \ge 1$ and $r \in \{0, 1\}$. Then $A$ starts with $+1$ and in the sequel negates all moves of $B$ until $B$ writes $\pm 2k$. Here "negates" means that $A$ writes $-(2j+1)$ or $(2j+1)$ according as $B$ writes $(2j)$ or $-(2j)$. We may assume that the first move of $B$ is $-2$ or else $A$ wins by writing $-3$ ($1+2-3=0$ is divisible by $N$ for all $N$). Moreover we may assume that $B$ also negates each move of $A$ up to step $2k$. If $B$ does not do so at step $2j$ then the expression ends $-(2j-2)+(2j-1)+(2j)$ or $+(2j-2)-(2j-1)-(2j)$ after step $2j$. So $A$ wins at step $2j+1$ by writing $-(2j+1)$ or $(2j+1)$ respectively, because $m-(m+1)-(m+2)+(m+3)=0$ for all $m$. Thus the sum $1-2+3-\cdots+(2k-1)-(2k)$ is obtained at step $2k$, after a move of $B$.

Neither player has won the game by that moment. Indeed denote $S_n = 1 - 2 + \cdots + (-1)^{n-1}n$, then $S_n = \frac{n+1}{2}$ for $n$ odd and $S_n = -\frac{n}{2}$ for $n$ even. It follows that if $1 \le i < j \le n$ then $|S_i - S_j| = \frac{j-i}{2}$ for $i, j$ of the same parity and $|S_i - S_j| = \frac{i+j+1}{2}$ for $i, j$ of different parity. In particular $0 < |S_i - S_j| \le n$. So there is no winner yet if $N > n$. This is the case here, with $N = 4k + r \ge 4k$ and $n = 2k$. Note that the argument works for $k=1$ ($N=4$) where $2k-2=0$ is not present in the sum but can be assumed.

Now $A$ wins by writing $-(2k+1)$. If $N = 4k + 1$ then the sum of the last two numbers $-(2k) - (2k+1) = -N$ is divisible by $N$; if $N = 4k$ then so is the sum of the last four numbers $-(2k-2) + (2k-1) - (2k) - (2k+1) = -4k = -N$.

Player $B$ has an analogous winning strategy for $N = 4k + r$ where $k \ge 1$ and $r \in \{2, 3\}$. Without loss of generality $A$ starts the game with $+1$ (if $B$ has a winning strategy for a game starting $+1$, he can just negate his moves in this strategy if $A$'s opening move is $-1$). Now $B$ answers $-2$ and in general negates $A$'s moves up to step $2k+1$. One may suppose again that $A$ also negates $B$'s moves, due to the identity $m - (m+1) - (m+2) + (m+3) = 0$. Thus the sum $1 - 2 + \cdots + (2k-1) - (2k) + (2k+1)$ is obtained at step $2k+1$, after a move of $A$. It was shown above that $0 < |S_i - S_j| \le 2k+1 < N$ whenever $1 \le i < j \le 2k+1$, so that there is no winner yet by that time. The next move $+(2k + 2)$ of $B$ is winning. If $N = 4k+3$ then $+(2k+1)+(2k+2) = N$; if $N = 4k+2$ then $+(2k-1)-(2k)+(2k+1)+(2k+2) = 4k+2 = N$.