Prove that because $a_{i} \in \mathbf{R}^{+}, i=1,2, \cdots, n, n \geqslant 2, n \in$ N, and $\sum_{i=1}^{n} a_{i}=1$, so
$$\begin{array}{l} \sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}} \\ =\frac{a_{1}^{2}}{2 a_{1}-a_{1}^{2}}+\frac{a_{2}^{2}}{2 a_{2}-a_{2}^{2}}+\cdots+\frac{a_{n}^{2}}{2 a_{n}-a_{n}^{2}} \\ \geqslant \frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)-\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)} \\ =\frac{1}{2-\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)} \end{array}$$

By the Cauchy-Schwarz inequality, we have $n\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right) \geqslant\left(a_{1}+\right.$
$$\left.a_{2}+\cdots+a_{n}\right)^{2}=1$$

Thus, $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant \frac{1}{n}$. Therefore, from (1) we get
$$\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}} \geqslant \frac{1}{2-\frac{1}{n}}=\frac{n}{2 n-1} .$$