$25$ checkers are placed on $25$ leftmost squares of $1 \times N$ board. Checker can either move to the empty adjacent square to its right or jump over adjacent right checker to the next square if it is empty. Moves to the left are not allowed. Find minimal $N$ such that all the checkers could be placed in the row of $25$ successive squares but in the reverse order.
1. **Initial Setup**: We have 25 checkers placed on the leftmost 25 squares of a \(1 \times N\) board. The checkers are numbered from 25 to 1 from left to right.

2. **Objective**: We need to find the minimal \(N\) such that all the checkers can be placed in a row of 25 successive squares but in the reverse order, i.e., checker #1 should be at the leftmost position and checker #25 at the rightmost position of these 25 squares.

3. **Checker #25's Final Position**: Checker #25 must jump over all 24 other checkers to reach its final position. Each jump moves the checker two squares to the right. Therefore, the minimum number of squares checker #25 must move is:
   \[
   2 \times 24 = 48
   \]
   Hence, checker #25's final position must be at least the 49th square.

4. **Checker #1's Movement**: If checker #25 is at the 49th position, checker #1 would not have moved. However, for the checkers to be in reverse order, checker #1 must move to the 25th position. This implies that checker #1 must move at least one square to the right during the process.

5. **Checker #2's Movement**: Similarly, checker #2 must also move at least one square to the right to allow the reordering process to continue. This is because no checker can jump over both checkers #1 and #2 if they are adjacent.

6. **Minimum \(N\)**: Given that checker #25 must be at least at the 49th position and checker #1 must move at least one square to the right, the minimal \(N\) must be:
   \[
   N \geq 50
   \]

7. **Verification**: To verify that \(N = 50\) is sufficient, we can outline a possible sequence of moves:
   - Move checker #1 one place to the right.
   - Then, in order, checkers #3, 5, 7, ..., 25, 24, 22, ..., 2 jump/move to their end positions, which is place number \(25 + i\) for checker #i.

Thus, the minimal \(N\) such that all the checkers can be placed in the row of 25 successive squares but in the reverse order is \(N = 50\).

The final answer is \(\boxed{50}\).