Let $n$ be a positive integer. Find all integers $k$ among $1,2,\dots,2n^2$ which satisfy the following condition:
・There is a $2n\times 2n$ grid of square unit cells. When $k$ different cells are painted black while the other cells are painted white, the minimum possible number of $2\times 2$ squares that contain both black and white cells is $2n-1$.
To solve the problem, we need to find all integers \( k \) among \( 1, 2, \dots, 2n^2 \) such that when \( k \) different cells are painted black in a \( 2n \times 2n \) grid, the minimum possible number of \( 2 \times 2 \) squares that contain both black and white cells is \( 2n-1 \).

1. **Define Cute Squares**:
   Let the \( 2 \times 2 \) squares that contain both black and white cells be called "cute" squares.

2. **Identify the Range of \( k \)**:
   We need to find the range of \( k \) such that the minimum number of cute squares is \( 2n-1 \).

3. **Case 1: \( 2n \mid k \)**:
   - If \( k \) is a multiple of \( 2n \), color the top \( \frac{k}{2n} \) rows black. This configuration ensures that there are exactly \( 2n-1 \) cute squares.
   - This is because each of the \( 2n-1 \) rows below the black rows will have exactly one \( 2 \times 2 \) square that contains both black and white cells.

4. **Case 2: \( n^2 - n + 1 \leq k \leq n^2 \)**:
   - Color an \( (n-1) \times n \) rectangle black such that one vertex of the rectangle meets the vertex of the grid.
   - Then place the remaining \( k - (n^2 - n) \) black cells right beneath the rectangle so that they are connected and meet the edge of the grid.
   - This configuration ensures that there are exactly \( 2n-1 \) cute squares.

5. **Case 3: \( k \geq n^2 - n + 1 \)**:
   - If \( k \leq n^2 - n \), there exists a rectangle \( k \times l \) such that \( kl \geq k \) and \( (k-1)l < k \), with \( k < 2n \) and \( l < 2n \).
   - Coloring black cells as in Case 2 gives a configuration with fewer than \( 2n-1 \) cute squares, which contradicts the problem's condition.

6. **Case 4: \( k > n^2 \)**:
   - If \( k > n^2 \), then \( 2n \) should divide \( k \).
   - If the rows have \( a_1, a_2, \ldots, a_{2n} \) black cells each, placing them in the first \( a_i \) cells gives the lowest number of cute squares.
   - Doing this on columns, we can say that \( a_i \geq a_{i+1} \) for all \( 1 \leq i \leq 2n-1 \).

7. **Subcases for \( k > n^2 \)**:
   - If \( a_1 < 2n \) and \( a_{2n} > 0 \), by previous analysis, there are at least \( \sum_{i=1}^{2n-1} (a_i - a_{i+1} + 1) = 2n-1 + a_1 - a_{2n} \), so it can only be \( 2n-1 \) when \( a_i \) is constant. This gives \( 2n \mid k \).
   - If \( a_1 = 2n \) and \( a_{2n} = 0 \), consider the least \( j \) such that \( a_{j+1} < 2n \). If \( a_{j+1} > 0 \), there are at least \( 2n - q + q = 2n \) cute squares, which is a contradiction.
   - If \( a_1 = 2n \) and \( a_{2n} > 0 \), consider the least \( j \) such that \( a_{j+1} < 2n \). There are at least \( 2n - a_{j+1} + \sum_{i=j+1}^{2n-1} (a_i - a_{i+1} + 1) = 4n - 1 - j - a_{2n} \) cute squares. Hence \( j + a_{2n} \geq 2n \), and \( k \geq j \times 2n + a_{2n} \times (2n - j) \geq j^2 - 2nj + 4n^2 \geq 3n^2 \), which is a contradiction.
   - If \( a_1 < 2n \) and \( a_{2n} = 0 \), consider the largest \( j \) such that \( a_j > 0 \). There are at least \( \sum_{i=1}^{j-1} (a_i - a_{i+1} + 1) + a_j = j - 1 + a_1 \) cute squares. Hence \( a_1 \leq 2n - j \), and \( k \leq a_1 \times j \leq j(2n - j) \leq n^2 \).

8. **Conclusion**:
   - If \( n^2 - n + 1 \leq k \leq n^2 \), there are at least \( 2n-1 \) cute squares.
   - The proof is the same as in the previous cases.

The final answer is \( \boxed{\forall k \mid n^2 - n + 1 \leq k \leq n^2 \text{ or } 2n \mid k, k \geq n^2 - n + 1} \)