Several  fleas sit on the squares of a $10\times 10$ chessboard (at most one  fea per square). Every minute, all  fleas simultaneously jump to adjacent squares. Each  fea begins jumping in one of four directions (up, down, left, right), and keeps jumping in this direction while it is possible; otherwise, it reverses direction on the opposite. It happened that during one hour, no two  fleas ever occupied the same square. Find the maximal possible number of  fleas on the board.
1. **Initial Observation**:
   - The chessboard is a $10 \times 10$ grid.
   - Fleas jump to adjacent squares in one of four directions (up, down, left, right) and reverse direction when they hit the edge.
   - No two fleas ever occupy the same square during one hour.

2. **Upper Bound**:
   - Each row and each column can have at most 2 fleas. This is because if there were more than 2 fleas in any row or column, they would eventually collide.
   - Since there are 10 rows and 10 columns, the maximum number of fleas is $2 \times 10 = 20$ for rows and $2 \times 10 = 20$ for columns.
   - Therefore, the total maximum number of fleas is $20 + 20 = 40$.

3. **Construction of Example**:
   - We need to construct an example where 20 fleas are on white squares and 20 fleas are on black squares, ensuring no two fleas ever occupy the same square.
   - The chessboard is colored in a checkerboard pattern, so half the squares are white and half are black.

4. **White Squares**:
   - Label the rows $0, 1, 2, \ldots, 9$ from top to bottom and the columns $0, 1, 2, \ldots, 9$ from left to right.
   - For each $0 \le i \le 9$, let $r_i$ denote the number of minutes it takes for the flea in the $i$-th row to reach the leftmost column (column $0$).
   - Define $c_i$ similarly as the number of minutes it takes for the flea in the $i$-th column to reach the uppermost row (row $0$).

5. **Modulo 18**:
   - The motion of the fleas is periodic with period 18.
   - The condition for the flea in the $i$-th row not intersecting the flea in the $j$-th column is:
     \[
     \{r_i \pm j\} \cap \{c_j \pm i\} = \emptyset
     \]
   - This simplifies to:
     \[
     \{r_i \pm i\} \cap \{c_j \pm j\} = \emptyset
     \]

6. **Construction of Sets**:
   - Construct the sets $\{r_i \pm i\}$ and $\{c_j \pm j\}$.
   - Example construction:
     \[
     (r_0, r_1, r_2, r_3, r_4, r_5, r_6, r_7, r_8, r_9) = (0, 1, 4, 3, 4, 13, 12, 13, 10, 9)
     \]
     \[
     (c_0, c_1, c_2, c_3, c_4, c_5, c_6, c_7, c_8, c_9) = (4, 11, 12, 7, 8, 17, 16, 3, 2, 13)
     \]

7. **Combining Fleas on Black Squares**:
   - An analogous argument for fleas on black squares (or simply reflecting the construction over an axis of symmetry) yields the desired 40 fleas.

\[
\blacksquare
\]

The final answer is $\boxed{40}$