Singapore Mathematical Olympiad (SMO)

The answer is $n = p + 1$. We first show that $n \le p + 1$. Since we are interested only in divisibility by $p$, we assume that $0 \le a_{ik} \le p - 1$. Note that each row and each column can have at most one zero. Also notice that we can scale each row or column by scalar not divisible by $p$ without affecting the given property of the table. Consider the first row, which must have at least $n-1$ non-zero values, say in the first $n-1$ columns. By scaling those columns by an appropriate scalar, we may assume that the first $n-1$ values in the first row are all 1's. Now in the second row, the first $n-1$ values must be distinct, hence $n-1 \le p$, as desired.

We now construct for $n = p + 1$. Take the following grid. For $2 \le i, k \le p + 1$,
$$a_{11} = 0,\ a_{1k} = 1,\ a_{i1} = 1,\ a_{ik} = k - i.$$
For a valid quadruple $i, j, k, l$ as described in the problem, it is simple to check it works when $i = 1$ or $k = 1$. When $i \ge 2$ and $k \ge 2$, the number $a_{ik}a_{jl} - a_{il}a_{jk}$ turns out to be $(i-j)(k-l) \ne 0 \pmod{p}$.

$$\begin{pmatrix} 0 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 2 & 3 & 4 \\ 1 & 4 & 0 & 1 & 2 & 3 \\ 1 & 3 & 4 & 0 & 1 & 2 \\ 1 & 2 & 3 & 4 & 0 & 1 \\ 1 & 1 & 2 & 3 & 4 & 0 \end{pmatrix}$$