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.

Look at it from another angle: given $a, b$ find $n$ for which the inequality holds.
For starters, try $a = -8b - 1$:
$$(n+1)^2(-8b-1) + (8n^2+4n)b = $$ $$-(n^2 + 2n + 1)(8b+1) + 8n^2b + 4nb =$$
$$ -n^2 - 2(6b -1)n - 8b - 1 > 0$$
or
$$n^2 + 2(6b -1)n + 8b+1 < 0$$
which means that any 
$$n \lt -6b + 1 + 2\sqrt{9b^2 - 5b}$$
is good. Since $b \lt 0$, the RHS grows indefinitely.
