In each cell of $n$x$n$ is a number so that in every $3$x$3$ subtable the sum of numbers is negative and the sum of all numbers is positive.

In each unit cell of $$n\times n$$ table we have a number so that in every $$3\times 3$$ subtable the sum of numbers is negative and the sum of all numbers is positive. For which $$n\geq 4$$ we can have such an arangment?

Clearly we don’t have such an arangemnet if $$3\mid n$$. Now suppose $$3\nmid n$$. Then if $$n=7$$ (or $$3$$ or $$10$$ or $$13$$) we see that configuration

$$\begin{array} {|r|r|r|r|r|r|r|} \hline 7 & -1 & -1& 7&-1&-1&7 \ \hline -1 & -1 & -1& -1&-1&-1&-1 \ \hline -1 & -1 & -1& -1&-1&-1&-1 \ \hline 7 & -1 & -1& 7&-1&-1&7 \ \hline -1 & -1 & -1& -1&-1&-1&-1 \ \hline -1 & -1 & -1& -1&-1&-1&-1 \ \hline 7 & -1 & -1& 7&-1&-1&7 \ \hline \end{array}$$

Works. So I tried to generalised this for arbitrary $$3n+1$$, instead of $$7$$ we put positive $$a$$ and instead of $$-1$$ we put negative $$b$$. So I have to prove that there are such $$a,b$$ that satisfies $$a+8b<0$$ and $$(n+1)^2a+(8n^2+4n)b>0$$ for arbitrary $$n$$, but I fail to do that. Any idea how to solve this.