8 Let $n \geqslant 2$ be a positive integer, and let $n$ positive numbers $v_{1}, v_{2}, \cdots, v_{n}$ satisfy the following two conditions:
(1) $v_{1}+v_{2}+\cdots+v_{n}=1$;
(2) $v_{1} \leqslant v_{2} \leqslant \cdots \leqslant v_{n} \leqslant 2 v_{1}$.

Find the maximum value of $v_{1}^{2}+v_{2}^{2}+\cdots+v_{n}^{2}$.
8. The condition can be generalized to: $v_{1}+v_{2}+\cdots+v_{n}=1, v_{1} \leqslant v_{2} \leqslant \cdots \leqslant v_{n} \leqslant r v_{1}$.

Below is the proof: $v_{1}^{2}+v_{2}^{2}+\cdots+v_{n}^{2} \leqslant \frac{(r+1)^{2}}{4 m}$, where $r>1, r \in \mathbf{R}^{+}$.
For any $j \in \mathbf{Z}^{+}, 1 \leqslant j \leqslant n$, by the condition, we have $\left(v_{j}-v_{1}\right)\left(r v_{1}-v_{j}\right) \geqslant 0$, hence $r v_{1} v_{j}-r v_{1}^{2}-v_{j}^{2}+v_{1} v_{j} \geqslant 0$.

Summing the above inequality from $j=1$ to $n$, we get $\sum_{j=1}^{n} v_{j}^{2} \leqslant(r+1) v_{1}-m v_{1}^{2}=$ $-m\left(v_{1}-\frac{r+1}{2 m}\right)^{2}+\frac{(r+1)^{2}}{4 m} \leqslant \frac{(r+1)^{2}}{4 m}$.