Let us consider two moves: $2k-1$ and $2k$. Suppose that before the first of them there was a layout of cards $a_1, a_2, \ldots, a_{2k-1}, a_{2k}, a_{2k+1}, \ldots, a_{100}$, 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: $100$, $98$, $96$, $\ldots$, $4$, $2$, $1$, $3$, $5$, $\ldots$, $97$, $99$. 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.