49th Mathematical Olympiad in Ukraine

The square with number «1» which is the nearest to the center is located on the longest diagonal. The distance to the border is $1004$, therefore the distance to the nearest square is $502$. Let us show that this distance is the shortest possible. By condition the distance between two squares is determined by the minimum difference between horizontal and vertical lines of squares' location. Let us assess the distance from the nearest number «1» to the center. Note that in upper and lower $502$ horizontals there are $2\cdot 502=1004$ numbers «1», because each row on the square has exactly one number «1». Analogously in $502$ left and right verticals there are no more than $1004$ numbers «1». Therefore there is at least one number «1» in the central square $1005\times 1005$. And the distance to each square of this "big" square to the center is not more than $502$. Our solution is complete.