Since $11^2 > 100$, the table cannot contain numbers divisible by squares of prime numbers $p$, with $p \ge 7$.
Suppose that there exists a number of the table (situated on line $\ell$ and column $c$) divisible by a prime $p \ge 17$. Then there exists one more number on line $\ell$ (and column $c' \ne c$) divisible by $p$. Also, there exists one more number on column $c$ (and line $\ell' \ne \ell$) divisible by $p$. Therefore, in the cell $(\ell', c')$ must be a number divisible by $p$. In consequence, in the table must appear a number $N \ge 7p > 100$ – impossible.

| 11 | $3 \cdot 11$ | $3^3$ | $5^2$ |
| --- | --- | --- | --- |
| $5 \cdot 11$ | $7 \cdot 11$ | $3 \cdot 5$ | $3 \cdot 7$ |
| $7^2$ | $3 \cdot 5^2$ | 13 | $3 \cdot 13$ |
| $3^2 \cdot 5$ | $3^2 \cdot 7$ | $5 \cdot 13$ | $7 \cdot 13$ |

It follows that the table cannot contain 2-digit numbers that are odd multiples of 17 – 3 numbers, of 19 – 3 numbers, of 23 – 2 numbers, of 29 – 2 numbers, of 31 – 2 numbers and of 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 – one number each; in total, 26 numbers. On the other hand, there are 45 odd 2-digit numbers, so the table can contain at most $45 - 26 = 19$ numbers. Hence $n \le 4$.
The table from above is an example for $n = 4$.