I claim that answer is all natural numbers $n$ where $n$ is even. Here is my Proof: Before we start to proof, name cells as $c_{ji}$ where $j$ shows the row number, $i$ shows the column number and call target color of cell, which means for whites- turning black and for blacks- turning white We will divide the problem into $2$ cases. Where $n$ is odd or even. For odd n Lemma 1. Corner colors (That is, $c_{1,1},c_{3n,1},c_{1,3n},c_{3n,3n}$) only can be changed by only 1 $2 \cdot 2$ square ( that is, the square is unique). Since we have to change the corner colors, we must have to use 2 steps to the corners. Note that after using steps on that square, we can't make another step on it hence it would change the color. Note that these squares changes neighbour of corner cells to the target colors too. Now look at the cell $c_{3,1}$. To change the color of it, we have 2 choices, Either we use square that $c_{2,1}$ is inside it too, or we will use square that $ c_{4,1}$ is inside it. For the first case, since $c_{2,1}$ was already in target color, it will be changed back and we will be forced to use corner square which is contraction to Lemma 1, hence we have to use second square with 2 steps. After that, with the same logic, we will look at $c_{5,1}$ and repeat the sequence, since this uses 2 cells, the sides must contain even cells, but $n$ is odd, hence contraction. For even n We can use the sequence above, and make the first row to the target color, noting that it makes the second row to the target color as well. After that, we will repeat the sequence to the 3th row. Since row has even numbers, we can turn all the rows to target colors

In order for the conditions in the problem to be fulfilled, it must be assumed that the number of squares in the table (which equates to 3n*3n, or 9n^2) must be divisible by 4, as the cells are inverted in color by multiples of 4 cells. As a result, n must be equal to the square root of (4/9)x, which, when n is substituted for a even positive integer, results in x equating to 9((n/2)^2)). Therefore, since when you substitute all even positive integers into the aforementioned 9((n/2)^2) you receive multiples of the coefficient of the area of the table (9, from 9n^2), it proves that the solution set for n is all positive even integers.