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.