Let $m,n$ be positive integers. Alice and Bob play a game on an initially blank $n\times n$ square grid $G$, alternating turns with Alice first. On a turn, the player can color a completely blank $m \times m$ subgrid of $G,$ or color in one blank cell of $G$. A player loses when they cannot do this. Find all $(m,n)$ so that with an optimal strategy, Alice wins.


1. **Case 1: \( m > n \)**
   - If \( m > n \), Alice can always color a single cell on her first move. Since the grid is \( n \times n \), there are \( n^2 \) cells in total.
   - If \( n \) is odd, the total number of cells is odd. Therefore, Alice, who starts first, will color the last cell and win.
   - If \( n \) is even, the total number of cells is even. Therefore, Bob, who plays second, will color the last cell and win.
   - Thus, Alice wins if and only if \( n \) is odd.

2. **Case 2: \( m \leq n \)**
   - We claim that Alice wins if and only if \( m \) is even or \( n \) is odd.
   
   - **Subcase 2.1: \( m \) and \( n \) have the same parity**
     - Suppose \( m \) and \( n \) are both odd or both even.
     - Alice can color an \( m \times m \) square that is centralized in the \( n \times n \) grid.
     - After Alice's move, Bob will make a move. Alice can then respond by making the same move reflected over the center of the \( n \times n \) grid.
     - Since Bob cannot color two squares that are reflections over the center in a single move, Alice will always have a move available.
     - Therefore, Alice wins by always mirroring Bob's moves.

   - **Subcase 2.2: \( n \) is odd and \( m \) is even**
     - Alice colors the center square of the grid on her first move.
     - After Bob's move, Alice can again mirror Bob's move over the center of the grid.
     - Since \( n \) is odd, the center square is uniquely defined, and Alice can always respond to Bob's move.
     - Therefore, Alice wins by always mirroring Bob's moves.

   - **Subcase 2.3: \( n \) is even and \( m \) is odd**
     - In this case, each move colors an odd number of squares.
     - Since the game ends when the entire board is colored, and the total number of squares \( n^2 \) is even, the game must end after an even number of moves.
     - Since Alice starts first, the game will end on Bob's turn.
     - Therefore, Bob wins regardless of the moves made.

Combining these points, we conclude that Alice wins if and only if \( m \) is even or \( n \) is odd.

The final answer is \( \boxed{ (m \text{ is even}) \text{ or } (n \text{ is odd}) } \).