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.

$-(m+4k+4)$	$-(m+4k+3)$	...	$-(m+2k+3)$	$m+1$
$-(m+2k+2)$				$m+2$
$-(m+2k+1)$		$(2k+1)\times (2k+1)$ table from the k-th step		...
...				$m+2k+1$
$-(m+2)$				$m+2k+2$
$-(m+1)$	$m+2k+3$	...	$m+4k+3$	$m+4k+4$

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. ■