8.35 How many natural numbers $n$ make $n^{2}-19 n+91$ a perfect square?
(China Beijing Junior High School Grade 2 Mathematics Competition, 1991)
[Solution] Since $n^{2}-19 n+91=(n-9)^{2}+(10-n)$.
When $n>10$,
$$
\begin{array}{l}
n^{2}-19 n+91=(n-9)^{2}+(10-n)
\end{array}
$$
is not a perfect square. Therefore, when $n>10$, there are no perfect squares.
Thus,
$$
1 \leqslant n \leqslant 10 .
$$

For $n=1,2,3,4,5,6,7,8,9,10$, verify one by one, only when $n=9, n=10$, $n^{2}-19 n+91$ is a perfect square.
Therefore, there are only 2 natural numbers $n$ that meet the requirement.