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 a $n\times n$ 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 arangement?


Clearly we don't have such an arangement if $3\mid n$. Now suppose $3\nmid n$. Then if $n=7$ (or $3$ or $10$ or $13$) we see that a 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 generalise this for an arbitrary $3n+1$, instead of $7$ we put a positive $a$ and instead of $-1$ we put a 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 an arbitrary $n$, but I fail to do that. Any idea how to solve this.

For every $n$ the quotient
$$
Q = \frac{8n^2 + 4n}{(n+1)^2} < 8.
$$
That means you can always find $a$ and $b$ such that
$$
Q < \frac{a}{b} < 8.
$$
$Q$ has limit $8$ so if you want integers for $a$ and $b$ they will have to be large when $n$ is large.
