7. If $n^{2}-19 n+91$ is a perfect square, find the value of the natural number $n$.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
$$
\begin{array}{l}
n^{2}-19 n+91=(n-9)^{2}+(10-n), \\
n^{2}-19 n+91=(n-10)^{2}+(n-9) .
\end{array}
$$

When $n>10$,
$$
(n-10)^{2}<n^{2}-19 n+91<(n-9)^{2} \text {, }
$$

a contradiction. Therefore, $n \leqslant 10$. Direct verification shows that $n=9,10$.