### Part (a)
1. Let $a$ be the number of fields labeled with $+1$ and $b$ be the number of fields labeled with $-1$. Then we have:
$$a + b = n^2$$
2. The sum $S_n$ of all the row and column sums is calculated. Each field in the grid is counted twice in $S_n$ (once in its row sum and once in its column sum). Therefore, we can express $S_n$ as:
$$S_n = 2a - 2b = 2(a - b)$$
3. For $S_n$ to be zero, we need:
$$2(a - b) = 0 \implies a - b = 0 \implies a = b$$
4. Since $n$ is odd, $n^2$ is also odd. Therefore, $a + b = n^2$ is odd. If $a = b$, then $a + b = 2a$ would be even, which contradicts the fact that $n^2$ is odd.
5. Hence, for no odd number $n$, there is a labeling such that $S_n = 0$.

$\blacksquare$

### Part (b)
1. If $n$ is even, then $n^2$ is also even. We need to show that there are at least six different labelings such that $S_n = 0$.
2. For $S_n = 0$, we need $a = b$. Since $n^2$ is even, we can have $a = b = \frac{n^2}{2}$.
3. The number of ways to choose $\frac{n^2}{2}$ fields out of $n^2$ fields to label with $-1$ is given by the binomial coefficient:
$$\binom{n^2}{\frac{n^2}{2}}$$
4. We need to show that this number is at least 6 for $n \geq 2$. For $n = 2$, we have:
$$\binom{4}{2} = 6$$
5. For $n > 2$, $\binom{n^2}{\frac{n^2}{2}}$ is even larger. Therefore, there are at least six different labelings such that $S_n = 0$.

$\blacksquare$