Consider a board of $a \times b$, with $a$ and $b$ integers greater than or equal to $2$. Initially their squares are colored black and white like a chess board. The permitted operation consists of choosing two squares with a common side and recoloring them as follows: a white square becomes black; a black box turns green; a green box turns white. Determine for which values of $a$ and $b$ it is possible, by a succession of allowed operations, to make all the squares that were initially white end black and all the squares that were initially black end white. 

Clarification: Initially there are no green squares, but they appear after the first operation.
**
   - For \(a\) and \(b\) both even, we can tile the board using \(2 \times 3\) rectangles. Each \(2 \times 3\) rectangle can be transformed to invert the colors.
   - For \(a\) even and \(b\) odd (or vice versa), one side must be a multiple of 3. We can use \(2 \times 3\) rectangles to cover the board and achieve the desired transformation.
   - For \(a\) and \(b\) both odd, the number of black squares and white squares cannot be equal modulo 3, making the transformation impossible.

5. **Conclusion:**
   The necessary and sufficient condition for the transformation to be possible is that \(6\) divides \(ab\). This ensures that the board can be tiled with \(2 \times 3\) rectangles, allowing the transformation.

\[
\boxed{6 \mid ab}
\]