4. 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
4.39.

Notice that
$$
\begin{array}{l}
\frac{1}{k \sqrt{k+1}+(k+1) \sqrt{k}}=\frac{(k+1) \sqrt{k}-k \sqrt{k+1}}{(k+1)^{2} k-k^{2}(k+1)} \\
=\frac{(k+1) \sqrt{k}-k \sqrt{k+1}}{k^{2}+k}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}} .
\end{array}
$$

Then the original expression $=1-\frac{1}{\sqrt{n+1}}$.
Given $\frac{19}{20}<1-\frac{1}{\sqrt{n+1}}<\frac{20}{21}$, we have
$$
20<\sqrt{n+1}<21 \Rightarrow 20^{2}-1<n<21^{2}-1 \text {. }
$$

Therefore, the maximum value of the positive integer $n$ is $21^{2}-2$ and the minimum value is $20^{2}$. The difference between them is $21^{2}-20^{2}-2=41-2=39$.