Example 2 Given
$$
\begin{array}{l}
\frac{1}{1 \times \sqrt{2}+2 \sqrt{1}}+\frac{1}{2 \sqrt{3}+3 \sqrt{2}}+\cdots+ \\
\frac{1}{n \sqrt{n+1}+(n+1) \sqrt{n}}
\end{array}
$$

is greater than $\frac{19}{20}$ and less than $\frac{20}{21}$. Then the difference between the maximum and minimum values of the positive integer $n$ is $\qquad$
(2009, I Love Mathematics Junior High School Summer Camp Mathematics Competition)
Consider the general term. The $k$-th term is
$$
\begin{array}{l}
\frac{1}{k \sqrt{k+1}+(k+1) \sqrt{k}} \\
=\frac{1}{\sqrt{k} \sqrt{k+1}(\sqrt{k}+\sqrt{k+1})} \\
=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k} \sqrt{k+1}} \\
=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}(k=1,2, \cdots, n) .
\end{array}
$$

Thus, the original expression is
$$
\begin{array}{l}
=\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\cdots+\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right) \\
=1-\frac{1}{\sqrt{n+1}} .
\end{array}
$$

From the given information,
$$
\begin{array}{l}
\frac{19}{20}\frac{1}{\sqrt{n+1}}>\frac{1}{21} \\
\Rightarrow 20<\sqrt{n+1}<21 \\
\Rightarrow 20^{2}-1<n<21^{2}-1 .
\end{array}
$$

Since $n$ is a positive integer,
$$
n_{\max }=21^{2}-2, n_{\text {min }}=20^{2} \text {. }
$$

Therefore, $21^{2}-2-20^{2}=39$ is the answer.