Answer: 100000.

Solution: Divide the board into $100^{2}$ squares of size $10 \times 10$. We will show that in each square, no more than 10 leopards can stand without attacking each other - from this it will follow that the total number of leopards cannot exceed $100^{2} \cdot 10=100000$.

Consider an arbitrary square $Q$ of size $10 \times 10$ and an arbitrary leopard $g$ in it. Leopard $g$ attacks all cells of the square, except for the cells that lie in the same row or column as it. If one of the other leopards $g^{\prime}$ in the square $Q$ stands in the same row as $g$, and another, $g^{\prime \prime}$, stands in the same column as $g$, then $g^{\prime}$ and $g^{\prime \prime}$ stand in different rows and columns and, consequently, attack each other; this is impossible. Otherwise, without loss of generality, all leopards in the square $Q$ stand in the same row as $g$, meaning there are no more than 10 of them.

Thus, we have shown that the total number of leopards cannot exceed 100,000; it remains to provide an example where this estimate is achieved. Number the columns of the board consecutively with the numbers $1,2, \ldots, 1000$. Place leopards on all cells of the columns whose numbers are divisible by 10. There will be $1000 \cdot 100=100000$ of these leopards, and they will not attack each other.

Comment: Only an example of placing 100000 leopards that do not attack each other is given - 2 points.

Examples with fewer leopards are not evaluated.

If only the proof that the number of leopards cannot exceed $100000$ is provided - 5 points.

If this estimate is only reduced to proving that in any $10 \times 10$ square there are no more than 10 leopards - 2 points instead of 5 for this part of the solution.
10.8. Given a $1000 \times 1000$ chessboard. A figure called a cheetah, from any cell $x$, attacks all cells of a $19 \times 19$ square centered at $x$, except for the cells that are in the same row or column as $x$. What is the maximum number of non-attacking cheetahs that can be placed on the board?

(I. Bogdanov)