a) Example: we number the rows and columns from $1$ to $9$. We write $1$ in the fields $(i, i)$ ($i = 1, \dots, 9$); $-1$ in field $(1, 9)$ and in fields $(i, i-1)$ ($i = 2, \dots, 9$); $0$ in other fields. Possible sums are $1$, $0$, and $-1$.

Evaluation: Suppose there are at least $19$ non-zero numbers. From Dirichlet's principle, there will be at least three non-zero numbers on any row, and therefore at least two non-zero numbers with the same sign, let's say positive ones (the situation with negative ones is analogous). Let's arrange the numbers in this order by size: $a_1 \le a_2 \le \dots \le a_9$, where $a_9 \ge a_8 > 0$. If $a_5 \ge 0$, then $a_5 + a_6 + a_7 + a_8 + a_9 \ge a_8 + a_9 > a_9$ must be of the same order, a contradiction. If $a_5 < 0$, then $a_1 + a_2 + a_3 + a_4 + a_5 < a_1$ must be of the same order, a contradiction.

b) A possible example without zeros is as follows (works because $5 \cdot 3 + 3 \cdot (-4) = 3$ and $4 \cdot 3 + 4 \cdot (-4) = -4$):

| 3 | 3 | 3 | 3 | 3 | -4 | -4 | -4 | -4 |
|----|----|----|----|----|----|----|----|----|
| -4 | 3 | 3 | 3 | 3 | 3 | -4 | -4 | -4 |
| -4 | -4 | 3 | 3 | 3 | 3 | 3 | -4 | -4 |
| -4 | -4 | -4 | 3 | 3 | 3 | 3 | 3 | -4 |
| -4 | -4 | -4 | -4 | 3 | 3 | 3 | 3 | 3 |
| 3 | -4 | -4 | -4 | -4 | 3 | 3 | 3 | 3 |
| 3 | 3 | -4 | -4 | -4 | -4 | 3 | 3 | 3 |
| 3 | 3 | 3 | -4 | -4 | -4 | -4 | 3 | 3 |
| 3 | 3 | 3 | 3 | 3 | -4 | -4 | -4 | -4 |