Example 7 Let $a_{i} \in \mathbf{R}^{+}, i=1,2, \cdots, n$, and $\sum_{i=1}^{n} a_{i}=1$, find
$$M=\sum_{i=1}^{n} \frac{a_{i}}{1+\sum_{j \neq i, j=1}^{n} a_{j}}$$

the minimum value.
Solve: 
\[
\begin{aligned}
M+n & =\left(\frac{a_{1}}{2-a_{1}}+1\right)+\left(\frac{a_{2}}{2-a_{2}}+1\right)+\cdots+\left(\frac{a_{n}}{2-a_{n}}+1\right) \\
& =\frac{2}{2-a_{1}}+\frac{2}{2-a_{2}}+\cdots+\frac{2}{2-a_{n}}\left(\text { by } H_{n} \leqslant A_{n}\right) \\
& \geqslant \frac{n^{2}}{\frac{1}{2}\left(2-a_{1}\right)+\frac{1}{2}\left(2-a_{2}\right)+\cdots+\frac{1}{2}\left(2-a_{n}\right)} \\
& =\frac{n^{2}}{\frac{1}{2}(2 n-1)}=\frac{2 n^{2}}{2 n-1} .
\end{aligned}
\]

Therefore, \( M \geqslant \frac{n}{2 n-1} \), with equality if and only if \( a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n} \). Thus, the minimum value of \( M \) is \( \frac{n}{2 n-1} \).