All the numbers $1,2,...,9$ are written in the cells of a $3\times 3$ table (exactly one number in a cell) . Per move it is allowed to choose an arbitrary $2\times2$ square of the table and either decrease by $1$ or increase by $1$ all four numbers of the square. After some number of such moves all numbers of the table become equal to some number $a$. Find all possible values of $a$.

I.Voronovich	
1. **Labeling and Initial Setup**:
   We are given a \(3 \times 3\) table with numbers \(1, 2, \ldots, 9\) placed in the cells. We can perform moves that increase or decrease all four numbers in any \(2 \times 2\) sub-square by 1. We need to determine the possible values of \(a\) such that all numbers in the table become equal to \(a\) after some number of moves.

2. **Coloring the Table**:
   We color the table in a checkerboard pattern:
   \[
   \begin{array}{|c|c|c|}
   \hline
   R & W & R \\
   \hline
   W & R & W \\
   \hline
   R & W & R \\
   \hline
   \end{array}
   \]
   Here, \(R\) represents red cells and \(W\) represents white cells.

3. **Invariant Analysis**:
   Consider the sum of the numbers in the red cells minus the sum of the numbers in the white cells:
   \[
   S = (a_1 + a_3 + a_7 + a_9) - (a_2 + a_4 + a_6 + a_8)
   \]
   When we perform a move on any \(2 \times 2\) sub-square, the sum of the numbers in the red cells and the sum of the numbers in the white cells each change by the same amount (either all increase by 1 or all decrease by 1). Therefore, \(S\) remains invariant.

4. **Initial Value of the Invariant**:
   Initially, the numbers in the table are \(1, 2, \ldots, 9\). Let's calculate the initial value of \(S\):
   \[
   S_{\text{initial}} = (1 + 3 + 7 + 9) - (2 + 4 + 6 + 8) = 20 - 20 = 0
   \]
   Thus, \(S\) is initially 0 and remains 0 after any number of moves.

5. **Final Configuration**:
   When all numbers in the table become equal to \(a\), the invariant \(S\) must still hold:
   \[
   S_{\text{final}} = (a + a + a + a) - (a + a + a + a) = 0
   \]
   This confirms that the invariant is preserved.

6. **Sum of All Numbers**:
   The sum of all numbers in the table initially is:
   \[
   1 + 2 + \cdots + 9 = \frac{9 \times 10}{2} = 45
   \]
   When all numbers are equal to \(a\), the sum of all numbers is:
   \[
   9a
   \]
   Therefore, we have:
   \[
   9a = 45 \implies a = 5
   \]

The final answer is \(\boxed{5}\)