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 $+(2k-1)-(2k)+(2k+1)+(2k+2)=4k+2=N$.