Example 11 Let natural numbers $a, b, c, d$ satisfy $\frac{a}{b}+\frac{c}{d}<1$, and $a+c=20$, find the maximum value of $\frac{a}{b}+\frac{c}{d}$.
Solution: First, fix $a$ and $c$, and assume $a \leqslant c$. Let $b=a+p, d=c+q (p, q \in \mathbb{N})$. From $1>\frac{a}{b}+\frac{c}{d}=\frac{a}{a+p}+\frac{c}{c+q}=\frac{a c+a q+a c+p c}{a c+a q+p q+p c}$, we know $a c < a c$, which implies $p q \geqslant a c+1$. Therefore,
$$
\frac{a}{a+p q}+\frac{c}{c+1} \leqslant \frac{a}{a+a c+1}+\frac{c}{c+1}.
$$

This proves that $\frac{a}{a+p}+\frac{c}{c+q} \leqslant \frac{a}{a c+a+1}+\frac{c}{c+1}$
$$
=1-\frac{1}{(a+a c+1)(c+1)}.
$$

Since $a+c=20$ and $a \leqslant c$, there are 10 pairs of $(a, c)$: $(1,19),(2,18), \cdots, (10,10)$. The corresponding values of $(a+a c+1)(c+1)$ are $420, 741, 990, 1173, 1296, 1365, 1386, 1365, 1308, 1221$, respectively, with 1386 being the largest. Therefore, the maximum value sought is $1-\frac{1}{1386}=\frac{1385}{1386}$ (or $\left.\frac{7}{99}+\frac{13}{14}\right)$.