Given a positive integer $k$, a pigeon and a seagull play a game on an $n\times n$ board. The pigeon goes first, and they take turns doing the operations. The pigeon will choose $m$ grids and lay an egg in each grid he chooses. The seagull will choose a $k\times k$ grids and eat all the eggs inside them. If at any point every grid in the $n\times n $ board has an egg in it, then the pigeon wins. Else, the seagull wins. For every integer $n\geq k$, find all $m$ such that the pigeon wins.

[i]Proposed by amano_hina[/i]
1. **Case 1: \( n = k \)**
   - If \( n = k \), the board is \( k \times k \). The pigeon wins if and only if every grid on the board has an egg. Since the seagull can eat all the eggs in a \( k \times k \) grid, the pigeon must lay at least \( n^2 = k^2 \) eggs to ensure that after the seagull's turn, the pigeon can still fill the board. Therefore, the pigeon wins if \( m \geq k^2 \).

2. **Case 2: \( n \geq k+1 \)**
   - **Subcase 2.1: \( m \leq k^2 - 1 \)**
     - If \( m \leq k^2 - 1 \), the seagull can always choose a fixed \( k \times k \) rectangle and eat all the eggs inside it. Since \( m < n^2 \), the pigeon cannot win in one move. To fill the \( k \times k \) gap created by the seagull, the pigeon needs at least \( k^2 \) eggs, which is impossible since \( m \leq k^2 - 1 \). Therefore, the seagull wins.

   - **Subcase 2.2: \( m \geq k^2 + 1 \)**
     - If \( m \geq k^2 + 1 \), consider the configuration where the number of eggs is maximized, say \( t \). If \( t = n^2 \), the pigeon wins. Otherwise, the seagull eats the eggs in a \( k \times k \) rectangle. If there is a grid not in this rectangle that does not contain an egg, the pigeon can lay an egg in each grid inside the \( k \times k \) rectangle and then lay an egg on the empty grid. This results in one more egg than before, contradicting the choice of configuration. Hence, all grids outside the \( k \times k \) rectangle contain at least one egg, and the pigeon can lay an egg in each grid inside the \( k \times k \) rectangle and win. Therefore, the pigeon wins if \( m \geq k^2 + 1 \).

3. **Case 3: \( m = k^2 \)**
   - **Subcase 3.1: \( n \leq 2k - 1 \)**
     - If \( n \leq 2k - 1 \), the pigeon wins. For \( n = 2k - 1 \), there is a unique grid (in the center of the board) that belongs to all \( k \times k \) rectangles. The pigeon colors this grid black and starts laying eggs in the grids but not in the black one. After each seagull's turn, at most \( k^2 - 1 \) grids are empty, and the pigeon can eventually fill all the grids except the black one. Then, the seagull removes eggs from at most \( k^2 - 1 \) grids, and the pigeon lays one egg in each of them and also in the black grid, winning the game.

   - **Subcase 3.2: \( n \geq 2k \)**
     - If \( n \geq 2k \), the pigeon loses. Suppose that after a pigeon's turn, there are \( a \leq n^2 - 1 \) grids that contain at least one egg. The seagull can remove the eggs from at least \( a/4 \) grids, and the pigeon can lay eggs in at most \( k^2 \) more grids. The new number of grids that contain at least one egg is \( b \leq a - a/4 + k^2 = 3/4a + k^2 \leq 3/4(n^2 - 1) + n^2/4 < n^2 \), so \( b \leq n^2 - 1 \). Inductively, the pigeon can never fill all the grids.

The final answer is: If \( \boxed{ k \leq n \leq 2k - 1 } \), the pigeon wins for \( m \geq k^2 \). If \( n \geq 2k \), the pigeon wins for \( m \geq k^2 + 1 \).