$9 \cdot 64$ Given a natural number $n$ greater than 1. Let $a, b, c, d$ be natural numbers satisfying
$$
\frac{a}{b}+\frac{c}{d}<1, a+c \leqslant n,
$$

Find the maximum value of $\frac{a}{b}+\frac{c}{d}$.
［Solution］First, let $a+c=n$. If $b, d \geqslant n+1$, then
$$
\frac{a}{b}+\frac{c}{d} \leqslant \frac{n}{n+1},
$$

and obviously $\frac{n}{n+1}0$, when $b>\bar{b}$, $f(b+1)-f(b)$ $<0$. Therefore, when $b$ equals the smallest integer not less than $b_{1}$, $f(b)$ is maximized. From $(2 n-1)^{2}$ $+12(n+1)=4 n^{2}+8 n+13$, we know
$$
2 n+2<\sqrt{(2 n-1)^{2}+12(n+1)}<2 n+3 \text {, }
$$

so
$$
\frac{2 n}{3}+\frac{1}{6}<\vec{b}<\frac{2 n}{3}+\frac{1}{3} .
$$

Thus, it is easy to prove that the smallest integer not less than $b_{1}$ is $\left[\frac{2 n}{3}+\frac{1}{6}\right]+1$.
We have thus proven that if $a+c=n, \frac{a}{b}+\frac{c}{d}<1$, the maximum value of $\frac{a}{b}+\frac{c}{d}$ is
$$
1-\frac{1}{\left[\frac{2 n}{3}+\frac{7}{6}\right]\left(\left[\frac{2 n}{3}+\frac{7}{6}\right]\left(n-\left[\frac{2 n}{3}+\frac{1}{6}\right]\right)+1\right)} .
$$

Since this number increases monotonically with $n$, it is also the maximum value of $\frac{a}{b}+\frac{c}{d}$ under the condition $a+c \leqslant n, \frac{a}{b}+\frac{c}{d}<1$.