What is the greatest amount of rooks that can be placed on an $n\times n$ board, such that each rooks beats an even number of rooks? A rook is considered to beat another rook, if they lie on one vertical or one horizontal line and no rooks are between them.

[I]Proposed by D. Karpov[/i]
To determine the greatest number of rooks that can be placed on an $n \times n$ board such that each rook beats an even number of other rooks, we need to carefully analyze the placement and interactions of the rooks.

1. **Initial Placement**:
   - Place a rook on every square of the $n \times n$ board. This gives us $n^2$ rooks initially.

2. **Analyzing Rook Interactions**:
   - A rook can beat another rook if they are in the same row or column with no other rooks between them.
   - On a fully filled $n \times n$ board, each rook in the interior (not on the edge) beats $2(n-1)$ rooks (since there are $n-1$ rooks in its row and $n-1$ rooks in its column, excluding itself).

3. **Edge and Corner Cases**:
   - Rooks on the edges (but not in the corners) beat $2(n-1) - 1$ rooks (since they miss one interaction compared to interior rooks).
   - Rooks in the corners beat $2(n-1) - 2$ rooks (since they miss two interactions compared to interior rooks).

4. **Ensuring Even Interactions**:
   - We need each rook to beat an even number of other rooks. This means we need to remove rooks that beat an odd number of other rooks.
   - Rooks on the edges (excluding corners) beat an odd number of rooks ($2(n-1) - 1$), so we remove these rooks.
   - Rooks in the corners beat an even number of rooks ($2(n-1) - 2$), so they can stay.

5. **Iterative Removal**:
   - After removing the edge rooks, we are left with a smaller $(n-2) \times (n-2)$ board.
   - Repeat the process of removing edge rooks until we are left with a board where all remaining rooks beat an even number of other rooks.

6. **Final Calculation**:
   - The process of removing edge rooks continues until we are left with a $2 \times 2$ board (if $n$ is even) or a $1 \times 1$ board (if $n$ is odd).
   - For an even $n$, the final board will be $2 \times 2$, and all 4 rooks will beat an even number of other rooks.
   - For an odd $n$, the final board will be $1 \times 1$, and the single rook will not beat any other rook (0 is even).

Therefore, the greatest number of rooks that can be placed on an $n \times n$ board such that each rook beats an even number of other rooks is:

\[
\boxed{4} \quad \text{if } n \text{ is even}
\]
\[
\boxed{1} \quad \text{if } n \text{ is odd}
\]