$4-$ $[\quad$ Evaluation + example $\quad]$

Car: Shapovosoov A.B.

On a strip of $1 \times N$, 25 checkers are placed on the leftmost 25 cells. A checker can move to the adjacent free cell to the right or jump over the adjacent checker to the right to the next cell (if this cell is free), movement to the left is not allowed. For what smallest $N$ can all the checkers be placed without gaps in reverse order?
From left to right, we number the cells and denote the checkers as $S_{1}, S_{2}, \ldots$

The checker $S_{25}$ cannot remain in place. Indeed, in this case, on the first move, $S_{24}$ will jump over it (there are no other moves) and will remain in place until the end (since at the end it must stand next to $S_{25}$). But then the other checkers will not be able to cross the barrier of two adjacent checkers. Contradiction.

Therefore, at the end, $S_{25}$ stands on cell 26 or further to the right, and the other 24 checkers stand to its right. Thus, $N \geq 50$.

We will show how to manage with 50 cells (in this case, the checker $S_{k}$ will occupy the (50-k)-th cell). First, $S_{25}$ moves to cell 26, then its place is taken by $S_{23}$ (jumping over $S_{24}$ and $S_{25}$ and stepping once to the right), then $S_{21}, \ldots$ Thus, all checkers with odd numbers occupy their places. Now we can sequentially send the checkers with even numbers to their places: $S_{2}$ (one step to the right and 23 jumps), $S_{4}, \ldots, S_{24}$.

## Answer

$N=50$.