Example 4 Let $a>b>c, n \in N^{+}$, and $\frac{1}{a-b}+\frac{1}{b-c}+$ $\frac{n}{c-a} \geqslant 0$ always holds, find the maximum value of $n$.

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Solution: Transposing terms, we get $\frac{n}{a-c} \leqslant \frac{1}{a-b}+\frac{1}{b-c}, n \leqslant(a-$
c) $\left(\frac{1}{a-b}+\frac{1}{b-c}\right), n \leqslant[(a-b)+(b-c)]\left(\frac{1}{a-b}\right.$ $\left.+\frac{1}{b-c}\right), n \leqslant(1+1)^{2}=4, \therefore$ the maximum value of $n$ is 4.