Squares of an $n \times n$ table ($n \ge 3$) are painted black and white as in a chessboard. A move allows one to choose any $2 \times 2$ square and change all of its squares to the opposite color. Find all such n that there is a finite number of the moves described after which all squares are the same color.
To solve the problem, we need to determine for which values of \( n \) it is possible to turn all squares of an \( n \times n \) chessboard into the same color using the given move. The move allows us to choose any \( 2 \times 2 \) square and change all of its squares to the opposite color.

1. **Case \( n \equiv 0 \pmod{4} \):**
   - **Claim:** If \( 4 \mid n \), then we can turn all cells of the grid into the same color.
   - **Proof:** It is sufficient to show that we can do this when \( n = 4 \), since we can divide larger squares into non-overlapping \( 4 \times 4 \) squares.
   - **Construction for \( n = 4 \):**
     - Start with the initial chessboard pattern:
       \[
       \begin{array}{cccc}
       B & W & B & W \\
       W & B & W & B \\
       B & W & B & W \\
       W & B & W & B \\
       \end{array}
       \]
     - Apply the move to the top-left \( 2 \times 2 \) square:
       \[
       \begin{array}{cccc}
       B & W & B & W \\
       W & B & W & B \\
       B & W & B & W \\
       W & B & W & B \\
       \end{array}
       \rightarrow
       \begin{array}{cccc}
       B & B & B & W \\
       B & B & W & B \\
       B & W & B & W \\
       W & B & W & B \\
       \end{array}
       \]
     - Continue applying the move to appropriate \( 2 \times 2 \) squares until all cells are the same color:
       \[
       \begin{array}{cccc}
       B & B & B & B \\
       B & B & B & B \\
       B & B & B & B \\
       B & B & B & B \\
       \end{array}
       \]
     - This shows that for \( n = 4 \), we can turn all cells into the same color. By extension, for any \( n \) divisible by 4, we can achieve the same result by dividing the board into \( 4 \times 4 \) squares and applying the same process to each.

2. **Case \( n \not\equiv 0 \pmod{4} \):**
   - **Claim:** If \( n \) is not divisible by 4, then we cannot turn all cells of the grid into the same color.
   - **Proof by induction:**
     - **Base case:** Consider \( n = 3 \). The initial pattern is:
       \[
       \begin{array}{ccc}
       B & W & B \\
       W & B & W \\
       B & W & B \\
       \end{array}
       \]
       - Any \( 2 \times 2 \) move will not be able to change the parity of the number of black and white cells in the grid. Since the number of black and white cells is not equal, it is impossible to make all cells the same color.
     - **Inductive step:** Assume the claim holds for \( n - 4 \). We need to show it holds for \( n \).
       - Consider \( n \equiv 1 \pmod{4} \). Without loss of generality, let the top-left cell be black. The parity of the number of black and white cells does not change after each operation. Since there are an even number of white cells, we must turn the whole grid black.
       - To turn the top row black, we must change the bolded rectangle:
         \[
         \begin{array}{cccc}
         B & \mathbf{W} & \mathbf{B} & W \\
         W & \mathbf{B} & \mathbf{W} & B \\
         \end{array}
         \]
       - Then we must change the next bolded rectangle:
         \[
         \begin{array}{cccc}
         B & B & \mathbf{W} & \mathbf{W} \\
         W & W & \mathbf{B} & \mathbf{B} \\
         \end{array}
         \]
       - Thus we obtain:
         \[
         \begin{array}{cccc}
         B & B & B & B \\
         W & W & W & W \\
         \end{array}
         \]
       - By induction, the claim is true. Since the second row has an odd number of white cells, we cannot turn all of them black. Thus, we cannot finish when \( n \equiv 1 \pmod{4} \).

3. **Conclusion:**
   - The other cases (\( n \equiv 2 \pmod{4} \) and \( n \equiv 3 \pmod{4} \)) follow similarly by reducing the case of \( n \) to the case of \( n - 4 \), yielding our desired result.

The final answer is \( \boxed{ n \equiv 0 \pmod{4} } \).