Let $n(n \geqslant 2)$ be a given positive integer, and real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1}+2 x_{2}+\cdots+n x_{n}=0$. Try to find the smallest real number $\lambda(n)$, such that
$$
\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \leqslant \lambda(n)\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right) \text {. }
$$
Three First, transform the condition into $\alpha\left(x_{1}+x_{2}+\cdots+x_{n}\right)=(\alpha-1) x_{1}+(\alpha-2) x_{2}+\cdots+(\alpha-n) x_{n}$, where $\alpha$ is a constant to be determined. Squaring both sides and applying the Cauchy-Schwarz inequality to the right side, we get
$$
\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \leqslant\left(\frac{n(n+1)(2 n+1)}{6 \alpha^{2}}-\frac{n(n+1)}{2 \alpha}+n\right)\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right),
$$

By completing the square, we find that the minimum value of $\frac{n(n+1)(2 n+1)}{6 \alpha^{2}}-\frac{n(n+1)}{2 \alpha}+n$ is $\frac{n^{2}-n}{4 n+2}$, and this minimum is achieved when $\alpha=\frac{2 n+1}{3}$.
From the equality condition of the Cauchy-Schwarz inequality, it is easy to see that when $x_{i}=\frac{2 n+1}{3}-i(i=1,2, \cdots, n)$, $\lambda(n) \geqslant \frac{n^{2}-n}{4 n+2}$.

Thus, the smallest $\lambda(n)$ is $\frac{n^{2}-n}{4 n+2}$.