Example 6 Arrange all the simplest fractions with denominators no greater than 99 in ascending order, find the two numbers adjacent to $\frac{17}{76}$.
Let $\frac{q}{p}$ and $\frac{n}{m}$ be the two numbers adjacent to $\frac{17}{76}$, and
$$\frac{q}{p}<\frac{17}{76}<\frac{n}{m}$$
with
$$\begin{array}{l}
17 p-76 q>0 \\
17 p-76 q \geqslant 1
\end{array}$$

First, consider the equation $17 p-76 q=1$ and find the largest positive integer solution for $p \leqslant 99$. To do this, we need to find a particular solution. Using
$$p=\frac{1}{17}(76 q+1)=4 q+\frac{1}{17}(8 q+1)$$

we can find a particular solution $(p, q)=(9,2)$. Thus, the general solution to this indeterminate equation is
$$(p, q)=(9+76 t, 2+17 t)$$
where $t$ is an integer. Under the condition $p \leqslant 99$, the maximum value of $p$ is 85, at which $q=19$.
On the other hand, from
$$\frac{17}{76}-\frac{q}{p}=\frac{17 p-76 q}{76 p}$$

we know that if $17 p-76 q \geqslant 2$, then
$$\frac{17}{76}-\frac{q}{p} \geqslant \frac{2}{76 p}=\frac{1}{38 p} \geqslant \frac{1}{38 \times 99}>\frac{1}{76 \times 85}=\frac{17}{76}-\frac{19}{85}$$

Therefore, under the given conditions, the number smaller than and closest to $\frac{17}{76}$ is $\frac{19}{85}$.
Similarly, we can determine that $\frac{n}{m}=\frac{15}{67}$.
In summary, the two numbers adjacent to $\frac{17}{76}$ in the sequence of these fractions are $\frac{19}{85}$ and $\frac{15}{67}$.