2005 IMO Shortlist S19

There are $n$ markers, each with one side white and the other side black. In the beginning, these $n$ markers are aligned in a row so that their white sides are all up. In each step, if possible, we choose a marker whose white side is up (but not one of the outermost markers), remove it, and reverse the closest marker to the left of it and also reverse the closest marker to the right of it. Prove that, by a finite sequence of such steps, one can achieve a state with only two markers remaining if and only if $n - 1$ is not divisible by $3$ .

*Proposed by Dusan Dukic, Serbia*

有 $n$ 标记，每个标记一侧为白色，另一侧为黑色。一开始，这些 $n$ 标记排成一行，使其白色面全部朝上。在每个步骤中，如果可能的话，我们选择一个白色面朝上的标记(但不是最外面的标记之一)，将其删除，然后反转其左侧最近的标记，并反转其右侧最近的标记。证明，通过这些步骤的有限序列，当且仅当 $n - 1$ 不能被 $3$ 整除时，我们可以实现仅剩下两个标记的状态。

*由塞尔维亚 Dusan Dukic 提议*