Consider a $L$-shaped piece formed by a square and two neighbouring squares.
The sum of its numbers is the opposite of the number that completes the $L$
to a $2 \times 2$ square, so the sum of numbers in any $L$ is less than $1$.
We proceed by induction on $n$. For $n=3$, divide the array in four regions: a $2 \times 2$ square, an $L$ and two single cells.

The sum is less than $3$, because of the single cells and the $L$.
For bigger odd $n$, divide the array in a square of side $n-2$, several squares of side $2$ and a $L$:

The sum of the square of side $n-2$ is less than $n-2$ by induction hypothesis and there are only an $L$ and a single square adding to less than $2$. So the sum is less than $n-2+2=n$ and the proof is complete.