Mathematical Olympiad Rioplatense

The second player $B$ has a winning strategy.

Each move does not affect (ignores) exactly two boxes $i,j$; then we denote it by $m_{i,j}$. Let $A$'s first move be $m_{1,2}$. Then $B$ divides the remaining boxes arbitrarily into $k-1$ pairs $\{3,4\},\{5,6\},\dots ,\{2k-1,2k\}$, and his first $k-1$ moves are $m_{3,4},m_{5,6},\dots ,m_{2k-1,2k}$. He is not interested in how $A$ plays until move $m_{2k-1,2k}$, the last one of these $k-1$. However $B$'s move after $m_{2k-1,2k}$ depends on $A$'s response. We show that this move of $B$, his $k$th, is winning.

Every box $i=1,2,\dots ,2k$ is ignored by exactly one of the moves $m_{1,2},m_{3,4},\dots ,m_{2k-1,2k}$. Hence these $k$ moves combined decrease the contents of every box by exactly $k-1$.

Let $m_{3,4}^{\prime },\dots ,m_{2k-1,2k}^{\prime }$ be $A$'s moves matching $m_{3,4},\dots ,m_{2k-1,2k}$. They affect a box at most $k-1$ times, so by the previous conclusion there will be at least $k-(k-1)=1$ pebbles in each box after $m_{2k-1,2k}^{\prime }$. In particular $m_{3,4}^{\prime },\dots ,m_{2k-1,2k}^{\prime }$ are not winning.

We claim that after $m_{2k-1,2k}^{\prime }$ at least two boxes contain exactly one pebble. A move ignores two boxes; then $k-1$ moves ignore at most $2k-2$ boxes. So there exist two boxes $i$ and $j$ that are affected by each of the moves $m_{3,4}^{\prime },\dots ,m_{2k-1,2k}^{\prime }$, meaning that the latter sequence decreases their contents by exactly $k-1$. But the previous sequence $m_{1,2},m_{3,4},\dots ,m_{2k-1,2k}$ decreased the contents of every box by exactly $k-1$. As a result each of $i$ and $j$ has exactly one pebble after $m_{2k-1,2k}^{\prime }$. It is clear now that $B$'s next move is winning.