Let $x_{0}, x_{1}, \cdots, x_{n}$ be positive integers, and $x_{0}<x_{1}$ $<\cdots<x_{n}, n \geqslant 1,[x, y]$ be the least common multiple of $x$ and $y$, and let
$$
S_{n}=\frac{1}{\left[x_{0}, x_{1}\right]}+\frac{1}{\left[x_{1}, x_{2}\right]}+\cdots+\frac{1}{\left[x_{n-1}, x_{n}\right]} \text {. }
$$

Try to find the maximum value of $S_{n}$.
When $n=1$, it is obvious that when $x_{0}=1, x_{1}=2$, the maximum value of $S_{1}$ is $\frac{1}{2}=1-\frac{1}{2^{2}}$;

When $n=2$, then when $x_{0}=1, x_{1}=2, x_{2}=2^{2}$, the maximum value of $S_{2}$ is $\frac{3}{4}=1-\frac{1}{2^{2}}$;
Continuing this approximation, we conjecture:
The maximum value of $S_{n}$ is $1-\frac{1}{2^{n}}$.
We will prove by mathematical induction that
$S_{n} \leqslant 1-\frac{1}{2^{n}}$:
First, when $n=1$,
Given $x_{1}>x_{0} \geqslant 1,\left[x_{0}, x_{1}\right] \geqslant 2$,
$\therefore S_{1}=\frac{1}{\left[x_{0}, x_{1}\right]} \leqslant \frac{1}{2}=1-\frac{1}{2}$.
The proposition holds.
Assume the proposition holds for $n=k$.
When $n=k+1$,
(1) If $x_{k+1} \geqslant 2^{k+1}$, then
$\frac{1}{\left[x_{k}, x_{k+1}\right]} \leqslant \frac{1}{x_{k+1}} \leqslant \frac{1}{2^{k+1}}$,
$\therefore S_{k+1} \leqslant 1-\frac{1}{2^{k}}+\frac{1}{2^{k+1}}=1-\frac{1}{2^{k+1}}$;
(2) If $x_{k+1}<2^{k+1}$, then
$$
\begin{array}{l}
\frac{1}{\left[x_{i}, x_{i+1}\right]}=\frac{\left(x_{i}, x_{i+1}\right)}{x_{i} x_{i+1}} \leqslant \frac{x_{i+1}-x_{i}}{x_{i} x_{i+1}} \\
=\frac{1}{x_{i}}-\frac{1}{x_{i+1}}(i=0,1, \cdots, k) .
\end{array}
$$

Where $\left(x_{i}, x_{i+1}\right)$ represents the greatest common divisor of $x_{i}, x_{i+1}$, and $\left(x_{i}, x_{i+1}\right) \leqslant\left|x_{i}-x_{i+1}\right|$ holds.
$$
\begin{aligned}
\therefore S_{k+1}= & \frac{1}{\left[x_{0}, x_{1}\right]}+\frac{1}{\left[x_{1}, x_{2}\right]}+\cdots \\
& +\frac{1}{\left[x_{k}, x_{k+1}\right]} \\
\leqslant & \left(\frac{1}{x_{0}}-\frac{1}{x_{1}}\right)+\left(\frac{1}{x_{1}}-\frac{1}{x_{2}}\right)+\cdots \\
& +\left(\frac{1}{x_{k}}-\frac{1}{x_{k+1}}\right) \\
= & \frac{1}{x_{0}}-\frac{1}{x_{k+1}} \\
\leqslant & 1-\frac{1}{2^{k+1}} .
\end{aligned}
$$

Combining (1) and (2), we see that the proposition also holds for $n=k+1$.

On the other hand, when $x_{0}=1, x_{1}=2, x_{2}=2^{2}, \cdots$, $x_{n}=2^{n}$, $S_{n}=1-\frac{1}{2^{n}}$.
Therefore, the maximum value of $S_{n}$ is $1-\frac{1}{2^{n}}$.