Suppose that all the cells adjacent to any particular cell must have different numerals. What is the minimum number of different numerals needed to fill a $5 \times 5$ square matrix?

Lower bound: at least $9$ numerals are necessary

Look at the central $3 \times 3$ block and its centre $e$ with neighbors

$\begin{pmatrix} a & b & c\\ d & e & f\\ g & h & i \end{pmatrix}$

By the rule at $e$, the eight neighbors $\{a,b,c,d,f,g,h,i\}$ must all be distinct (already $8$ different numerals).

Now suppose $e$ shared a numeral with any neighbor, say $e=a$. Consider the cell $b$: its neighbor set includes both $a$ and $e$, which would then have the same numeral — violating the requirement at $b$ that its neighbors be pairwise distinct. The same argument works for $e$ with any of the eight neighbors (there is always a cell adjacent to both).

Hence $e$ must also be different from each of its eight neighbors, forcing $9$ distinct numerals in this $3 \times 3$ block. Therefore the whole $5 \times 5$ grid needs at least $9$ numerals.

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Upper bound: $9$ numerals suffice (explicit construction)

Color cell $(r,c)$ by the pair $(r \bmod 3, c \bmod 3)$ mapped to the numerals $\{1,2,\dots,9\}$. Concretely, use the repeating $3 \times 3$ pattern

$\begin{pmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{pmatrix}$

tiled across the $5 \times 5$. Every $3 \times 3$ block then contains all nine distinct numerals, so in particular the $8$ neighbors around any centre are pairwise distinct. Edge and corner cells have fewer neighbors, which are a subset of some $3 \times 3$ block and thus still pairwise distinct.

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Conclusion

Both bounds match, so the minimum number of different numerals required is $\boxed{9}$.