31. Since
$$\begin{array}{c} \frac{x_{i}}{1+(n-1) x_{i}}=\frac{1}{n-1} \cdot \frac{1+(n-1) x_{i}-1}{1+(n-1) x_{i}}= \\ \frac{1}{n-1}-\frac{1}{n-1} \cdot \frac{1}{1+(n-1) x_{i}} \end{array}$$

Thus,
$$\begin{array}{l} \frac{x_{1}}{1+(n-1) x_{1}}+\frac{x_{2}}{1+(n-1) x_{2}}+\cdots+\frac{x_{n}}{1+(n-1) x_{n}} \leqslant 1 \Leftrightarrow \\ \frac{1}{1+(n-1) x_{1}}+\frac{1}{1+(n-1) x_{2}}+\cdots+\frac{1}{1+(n-1) x_{n}} \geqslant 1 \end{array}$$

By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l} {\left[\frac{1}{1+(n-1) x_{1}}+\frac{1}{1+(n-1) x_{2}}+\cdots+\frac{1}{1+(n-1) x_{n}}\right] \cdot} \\ \left.\left[1+(n-1) x_{1}\right]+\left[1+(n-1) x_{2}\right]+\cdots+\left[1+(n-1) x_{n}\right]\right\} \geqslant n^{2} \end{array}$$

And since $x_{1}+x_{2}+\cdots+x_{n} \leqslant n$, we have
$$\begin{array}{l} {\left[1+(n-1) x_{1}\right]+\left[1+(n-1) x_{2}\right]+\cdots+\left[1+(n-1) x_{n}\right]=} \\ n+(n-1)\left(x_{1}+x_{2}+\cdots+x_{n}\right) \leqslant n+\left(n^{2}-1\right) n=n^{2} \end{array}$$

Therefore:
$$\frac{1}{1+(n-1) x_{1}}+\frac{1}{1+(n-1) x_{2}}+\cdots+\frac{1}{1+(n-1) x_{n}} \geqslant 1$$