8 . The smallest positive integer $n$ that makes $n^{2}+2023 n$ a perfect square is $\qquad$
Solve:
Let $n^{2}+2023 n=m^{2} \Rightarrow 4 n^{2}-4 m^{2}+2023 \cdot 4 n+2023^{2}=2023^{2}$
$$
\Rightarrow(2 n+2 m+2023)(2 n-2 m+2023)=2023^{2}=7^{2} \times 17^{4} \text {, }
$$

Let $x=2 n+2 m+2023, y=2 n-2 m+2023$, from $x+y=4 n+2 \times 2023, (x+y)^{2}-(x-y)^{2}=4 x y$ we know, $x+y$ takes the minimum value if and only if $x-y$ is the smallest.
The positive divisors of $7^{2} \times 17^{4}$ in ascending order are $7,17,7^{2}, 7 \times 17,17^{2}, 7^{2} \times 17,7 \times 17^{2}, 17^{3}$,
$$
7^{2} \times 17^{2}, 7 \times 17^{3}, 17^{4}, 7^{2} \times 17^{3}, 7 \times 17^{4} \text {. }
$$

Therefore, $x+y=4 n+2 \times 2023 \geqslant 7^{2} \times 17+17^{3} \Rightarrow 4 n \geqslant 7^{2} \times 17+17^{3}-2 \times 7 \times 17^{2}$
$$
=7^{2} \times 17+3 \times 17^{2}=100 \times 17 \Rightarrow n \geqslant 25 \times 17=425 \text {. }
$$

Upon verification, $425^{2}+2023 \times 425=25^{2} \times 17^{2}+7 \times 17^{2} \times 25 \times 17$
$$
=25 \times 17^{2} \times(25+7 \times 17)=(5 \times 17 \times 12)^{2}=1020^{2} .
$$