The problem holds for and only for $n$ odd.

To this end, suppose $n$ even. Colour exactly one point in white and the rest in black. Choose $k=2$. After performing a 2-move, the number of points of a same colour remains odd, so a monochromatic configuration is not achievable.

Suppose $n$ odd. Assume $k$ odd. Consider an initial colouring with $p$ white points and $n-p$ black points. Let $p = ks+r$, $0 \le r < k$. Perform $s$ $k$-moves upon the white points to reach a configuration of $r$ white and $n-r$ black points. Perform a $k$-move with the $r$ white points and other $k-r$ black to obtain exactly $k-r$ white points. Therefore, one can obtain exactly $r$ or exactly $k-r$ white points, with only one of the numbers $r$ or $k-r$ being even, say $r$.

Perform a $k$-move with the $\frac{r}{2}$ white points to get exactly $k$ white points. After another $k$-move with the white points, all points turn black and we are done.

Assume $k$ even. Again, consider an initial colouring with $p$ white points and $n-p$ black points. One of the numbers $p$ or $n-p$ is even, say $p$. In $p = ks+r$, $0 \le r < k$, notice that $r$ is even. From this point on, proceed as in the previous case.