Four, (30 points) Let natural numbers $a, b, c, d$ satisfy $\frac{a}{b}+\frac{c}{d}<1$ and $a+c=20$. Find the maximum value of $\frac{a}{b}+\frac{c}{d}$.


Given $a, c$, and without loss of generality, assume $a \leqslant c$. Let $b=a+p, d=c+q$, where $p, q \in \mathbb{N}$. We have
$$
\begin{array}{l}
1>\frac{a}{b}+\frac{c}{d}=\frac{a}{a+p}+\frac{c}{c+q} \\
=\frac{a c+a q+a c+c p}{a c+a q+p q+c p} .
\end{array}
$$

This implies $a c < a c$. Therefore, $p q \geqslant a c+1$,
$$
\frac{a}{a+p q}+\frac{c}{c+1} \leqslant \frac{a}{a+a c+1}+\frac{c}{c+1} \text { . }
$$

Thus, we have
$$
\begin{aligned}
\frac{a}{a+p}+\frac{c}{c+q} & \leqslant \frac{a}{a+a c+1}+\frac{c}{c+1} \\
& =1-\frac{1}{(a+a c+1)(c+1)} .
\end{aligned}
$$

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