Example 6 Given pairwise distinct positive integers $a, b, c, d, e, f, g, h, n$ satisfying
$$n=a b+c d=e f+g h$$

Find the minimum value of $n$.
If none of $a, b, c, d, e, f, g, h$ equals 1, then
$$\begin{aligned}
2 n & =a b+c d+e f+g h \\
& \geqslant 4 \sqrt[4]{a b c d e f g h} \\
& \geqslant 4 \sqrt[4]{2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9} \\
& =4 \sqrt[4]{2^{7} \times 3^{4} \times 5 \times 7} \\
& =4 \times 4 \times 3 \times \sqrt[4]{\frac{35}{2}} \\
& >4 \times 4 \times 3 \times 2=96
\end{aligned}$$

Thus, $n \geqslant 48$. Suppose one of $a, b, c, d, e, f, g, h$ equals 1, without loss of generality, let $h=1$, then $2 n=a b+c d+e f+g$, and there exists a minimum value. In this case, $g$ must be the largest of these numbers. Therefore,
$$\begin{aligned}
2 n & =a b+c d+e f+g \\
& \geqslant g+3 \sqrt[3]{a b c d e f} \\
& \geqslant g+3 \sqrt[3]{2 \times 3 \times 4 \times 5 \times 6 \times 7} \\
& =g+3 \sqrt[3]{5040} \\
& >g+3 \sqrt[3]{4913}=g+51
\end{aligned}$$

Since $g \geqslant 8$, we have $2 n \geqslant 60, n \geqslant 30$. If $g \geqslant 9$, then $2 n \geqslant 61, n \geqslant 31$. Assume $n=30$, then $g=8$. Thus, $a, b, c, d, e, f, g, h$ is a permutation of the set $\{1,2,3,4,5,6,7,8\}$. In particular, one of the numbers is 5, say $a=5$. So $30=a b+c d$, which means $c d$ is divisible by 5. This is a contradiction.

Therefore, $n \geqslant 31$. Since $31=1 \times 7+4 \times 6=2 \times 8+3 \times 5$, the minimum value of $n$ is 31.