On $5\times 5$ squares, we cover the area with several S-Tetrominos (=Z-Tetrominos) along the square so that in every square, there are two or fewer tiles covering that (tiles can be overlap). Find the maximum possible number of squares covered by at least one tile.
1. **Understanding the Problem:**
   We need to cover a $5 \times 5$ grid using S-Tetrominos (Z-Tetrominos) such that each square is covered by at most two tiles. We aim to find the maximum number of squares that can be covered by at least one tile.

2. **Coloring the Grid:**
   Consider a $5 \times 5$ grid. We can color the grid in a checkerboard pattern with black and white squares. Each S-Tetromino covers exactly four squares, and we need to ensure that no square is covered by more than two tiles.

3. **Counting the Black Squares:**
   Let's color the grid such that there are 12 black squares and 13 white squares. Each S-Tetromino will cover exactly one black square. Since there are 12 black squares, we can place at most 12 S-Tetrominos.

4. **Placing the S-Tetrominos:**
   Each S-Tetromino covers 4 squares. If we place 12 S-Tetrominos, they will cover $12 \times 4 = 48$ squares. However, since each square can be covered by at most two tiles, we need to ensure that no square is counted more than twice.

5. **Ensuring No Overlap:**
   To maximize the number of squares covered, we need to ensure that the overlap is minimized. If we place the S-Tetrominos such that each square is covered exactly once, we can cover 24 squares without any overlap.

6. **Example Configuration:**
   Consider the following configuration where each S-Tetromino is placed without overlapping:
   ```
   S-Tetromino 1: (0,0), (0,1), (1,1), (1,2)
   S-Tetromino 2: (0,2), (0,3), (1,3), (1,4)
   S-Tetromino 3: (2,0), (2,1), (3,1), (3,2)
   S-Tetromino 4: (2,2), (2,3), (3,3), (3,4)
   S-Tetromino 5: (4,0), (4,1), (5,1), (5,2)
   S-Tetromino 6: (4,2), (4,3), (5,3), (5,4)
   ```

7. **Conclusion:**
   By placing the S-Tetrominos in such a way, we can cover 24 squares without any overlap. Therefore, the maximum number of squares that can be covered by at least one tile is 24.

The final answer is $\boxed{24}$