a) Let $a$ be the number of fields labeled with $1$ and $b$ be the number of fields labeled $-1$. Then we have $a+b=n^{2}$ and since every field is counted twice in $S_{n}$, we get $S_{n}=2a-2b=2(a-b)$. However, since $n$ is odd, $a=b$ is impossible and so $S_{n}=0$ cannot happen. b) We know from a) that it suffices to have an equal amount of $1$ and $-1$. There are $n^{2} \choose \frac{n^{2}}{2}$ possibilities to choose the fields labeled with $-1$. Each of these possibilities lead to a different label, and since $n^{2} \choose \frac{n^{2}}{2}$ $\geq 6$ for $n\geq 2$, we are finished.

redacted