62nd Ukrainian National Mathematical Olympiad

Let us consider two moves: $2k-1$ and $2k$. Suppose that before the first of them there was a layout of cards $a_1,a_2,\dots ,a_{2k-1},a_{2k},a_{2k+1},\dots ,a_{100}$, then after two moves we will have the following changes: $$a_{2k-1},a_{2k-2},\dots ,a_1,a_{2k},a_{2k+1},\dots ,a_{100}\to a_{2k},a_1,a_2,\dots ,a_{2k-1},a_{2k+1},\dots ,a_{100}.$$ Thus, after each pair of moves, the corresponding even number is moved to the first place. Therefore, after 100 moves, we will have such a layout of cards: $100$, $98$, $96$, $\dots$, $4$, $2$, $1$, $3$, $5$, $\dots$, $97$, $99$. We will call this perturbation a megamove. We will call the initial position $A_0$, the position after the first megamove $A_1$, after the second $A_2$, and so on. It is not difficult to understand that if the position $A_m$ arises from the position $A_{m+1}$, then vice versa, it cannot arise from any other position except for $A_m$. Therefore, let us consider the situation after a sufficiently large number of megamoves. They have to start repeating because there is a finite number of them possible. But if the position is repeated, then all previous ones coincide. Thus, the initial position had to repeat as well.