Iranian Mathematical Olympiad

The answer is yes. First, we choose a cell as origin and fill it with zero, then we fill the table in a way that the numbers in the cells of $(2k+1) \times (2k+1)$ table centered at the origin be a permutation of numbers
$$(-2k^2 - 2k, -2k^2 - 2k + 1, \dots, 0, \dots, 2k^2 + 2k - 1, 2k^2 + 2k),$$
We proceed the filling of the table inductively. Suppose that in step $k \ge 0$, the $(2k+1) \times (2k+1)$ table centered at the origin is filled with the before-mentioned numbers such that the sequence of numbers in each row and in each column is increasing. Putting $m = 2k^2+2k$. Now, consider $(2k+3) \times (2k+3)$ table centered at the origin. Fill this table as follows.

It can be easily verified that the columns and rows of this table form increasing sequences. So it is clear that with this algorithm the table will be filled with a permutation of integers and has the desired properties. ■