Example 6 Let positive integers $a, b, c, d$ satisfy $\frac{a}{b}+\frac{c}{d} < 1$ and $a+c=20$. Find the maximum value of $\frac{a}{b}+\frac{c}{d}$.
Solution: By symmetry, we may assume $a \leqslant c$. From the given conditions, we know $b>a, d>c$, so we can let
$b=a+p, d=c+q$ (where $p, q$ are positive integers).
From $1-\left(\frac{a}{b}+\frac{c}{d}\right)=1-\left(\frac{a}{a+p}+\frac{c}{c+q}\right)$
$$
=\frac{p q-a c}{(a+p)(c+q)}>0,
$$

we get $p q>a c$, thus, $p q \geqslant a c+1$.
Since $(a+p)(c+q)=a c+a q+p c+p q$
$$
\begin{array}{l}
=a c+p q+c p q+a-(q-1)(c p-a) \\
\leqslant a c+p q+c p q+a \\
=(a+p q)(c+1),
\end{array}
$$

therefore, $1-\left(\frac{a}{a+p}+\frac{c}{c+q}\right)$
$$
\begin{array}{l}
\geqslant \frac{p q-a c}{(a+p q)(c+1)} \\
=1-\left(\frac{a}{a+p q}+\frac{c}{c+1}\right) .
\end{array}
$$

Thus, $\frac{a}{b}+\frac{c}{d} \leqslant \frac{a}{a+p q}+\frac{c}{c+1}$
$$
\begin{array}{l}
\leqslant \frac{a}{a+a c+1}+\frac{c}{c+1} \\
=1-\frac{1}{(a+a c+1)(c+1)} .
\end{array}
$$

Equality in (1) holds when $q=1$ and $p q=a c+1$.
Therefore, the maximum value of $\frac{a}{b}+\frac{c}{d}$ is achieved when $(a+a c+1)(c+1)$ is maximized.

Since $a+c=20$ and $a \leqslant c$, there are 10 pairs of $(a, c)$ that satisfy the conditions, namely
$$
(1,19),(2,18), \cdots,(10,10),
$$

and the corresponding values of $(a+a c+1)(c+1)$ are 420, 741, 990, 1173, 1296, 1365, 1386, 1365, 1308, 1221, with 1386 being the largest.
Therefore, the maximum value sought is
$$
1-\frac{1}{1386}=\frac{1385}{1386}\left(\text { or } \frac{7}{99}+\frac{13}{14}\right).
$$