12 For any positive numbers $a_{1}, a_{2}, \cdots, a_{n}, n \geqslant 2$, find the minimum value of $\sum_{i=1}^{n} \frac{a_{i}}{S-a_{i}}$, where $S=$ $\sum_{i=1}^{n} a_{i}$.
12. Let $b_{i}=S-a_{i}$, then $\sum_{i=1}^{n} b_{i}=(n-1) S$. By the AM-GM inequality, we get $\sum_{i=1}^{n} \frac{a_{i}}{S-a_{i}}$
$$=\sum_{i=1}^{n}\left(\frac{a_{i}}{b_{i}}+1\right)-n=S \sum_{i=1}^{n} \frac{1}{b_{i}}-n \geqslant \frac{n S}{\sqrt[n]{b_{1} \cdots b_{n}}}-n \geqslant \frac{n^{2} S}{b_{1}+\cdots+b_{n}}-n=\frac{n}{n-1} .$$

When $a_{1}=a_{2}=\cdots=a_{n}=1$, the equality holds, hence the minimum value is $\frac{n}{n-1}$.