Example 1 (Question from the 11th "Hope Cup" Invitational Competition) Let $a > b > c, n \in \mathbf{N}$, and $\frac{1}{a-b}+\frac{1}{b-c} \geqslant \frac{n}{a-c}$ always holds, then the maximum value of $n$ is ( ).
A. 2
B. 3
C. 4
D. 5
Solution: Choose C. Reason: $\frac{1}{a-b}+\frac{1}{b-c} \geqslant \frac{n}{a-c} \Leftrightarrow \frac{a-c}{a-b}+\frac{a-c}{b-c} \geqslant n$, thus $n \leqslant\left[\frac{a-c}{a-b}+\frac{a-c}{b-c}\right]_{\text {min }}$. And $\frac{a-c}{a-b}+\frac{a-c}{b-c}=2+\frac{b-c}{a-b}+\frac{a-b}{b-c} \geqslant 4$, and when $2 b=a+c$, the value of the left side of the inequality equals 4. Therefore, $\left[\frac{a-c}{a-b}+\frac{a-c}{b-c}\right]_{\min }=4$.
Hence, $n \leqslant 4$, which means the maximum value of $n$ is 4.