Let $\displaystyle m$ the minimum such $\displaystyle n$. I came up with the following upper bound: $\displaystyle m \leq 26$ with the following solution to the puzzle

$\displaystyle \begin{bmatrix}3 & 6 & 15 & 25 & 5 \\24 & 9 & 18 & 20 & 10\\21 & 12 & 26 & 8 & 4\\7 & 14 & 22 & 16 & 2\end{bmatrix}$

If we inspect this table, we see that it satisfies the conditions for divisibility and the upper bound is correct. Now, if a prime $\displaystyle p$ appears explicitly in the table, its neighbors must be divisible by that prime. If the prime appears in the interior, its neighbors must be at least the sizes $\displaystyle 2p, 3p, 4p, 5p$. If it appears on an edge, the neighbors are at least $\displaystyle 2p, 3p, 4p$. If it appears in a corner the neighbors are at least $\displaystyle 2p, 3p$. Using the lowest of these upper bounds $\displaystyle 3p$, in an optimal solution we must have that $\displaystyle 3p \leq 25$, or $\displaystyle p \leq 26/3 \leq 9$. That is, the only explicit primes in this table are at most $\displaystyle 2,3,5$ and $\displaystyle 7$. No other primes can appear explicitly in the table. As we can see, up to 26, the only numbers that do not appear in this table are $\displaystyle 1, 11, 13, 17, 19$ and $\displaystyle 23$. Since these numbers are not allowed in an optimal solution, the solution I provided must be optimal. Therefore $\displaystyle m = 26$.