$n$ is a natural number. Given  $3n \cdot 3n$ table, the unit cells are  colored white and black such that starting from the left up corner diagonals are colored in pure white or black in ratio of 2:1 respectively. ( See the picture below). In one step any chosen $2 \cdot 2$ square's white cells are colored orange, orange are colored black and black are colored white. Find all $n$  such that with finite steps, all the white cells in the table turns to black, and all black cells in the table turns to white. ( From starting point)
1. **Define the Problem and Notation:**
   - We have a \(3n \times 3n\) table where cells are colored in a specific pattern: starting from the top-left corner, diagonals are colored in pure white or black in a ratio of 2:1 respectively.
   - We can choose any \(2 \times 2\) square and change the colors of its cells: white to orange, orange to black, and black to white.
   - We need to determine for which natural numbers \(n\) it is possible to turn all white cells to black and all black cells to white in a finite number of steps.

2. **Initial Observations:**
   - The table is divided into \(3n\) rows and \(3n\) columns.
   - The coloring pattern follows a diagonal rule with a 2:1 ratio of white to black cells.

3. **Case 1: \(n\) is odd**
   - **Lemma 1:** Corner colors (i.e., \(c_{1,1}\), \(c_{3n,1}\), \(c_{1,3n}\), \(c_{3n,3n}\)) can only be changed by a unique \(2 \times 2\) square.
   - To change the color of a corner cell, we must use exactly one \(2 \times 2\) square that includes the corner cell.
   - After using this square, we cannot use it again without reverting the color change.
   - Consider the cell \(c_{3,1}\). To change its color, we have two choices:
     - Use a square that includes \(c_{2,1}\), which would revert \(c_{2,1}\) back to its original color, forcing us to use the corner square again, contradicting Lemma 1.
     - Use a square that includes \(c_{4,1}\), which requires two steps.
   - This logic extends to all cells in the first column, requiring an even number of cells in each row and column to achieve the target color.
   - Since \(n\) is odd, the number of cells in each row and column is odd, leading to a contradiction.

4. **Case 2: \(n\) is even**
   - We can use the sequence described above to change the colors of the first row to the target color.
   - This process will also change the second row to the target color.
   - We can repeat this sequence for the third row and so on.
   - Since \(n\) is even, the number of cells in each row and column is even, allowing us to turn all rows to the target colors.

Conclusion:
- For \(n\) to be such that all white cells can be turned to black and all black cells to white in a finite number of steps, \(n\) must be even.

The final answer is \( \boxed{ n } \) is even.