One, (40 points) Let $x_{1}, x_{2}, \cdots, x_{n}$ be real numbers no less than -1, satisfying $\sum_{i=1}^{n} x_{i}^{3}=0$. Find the smallest constant $c$, such that for all positive integers $n$, we have $\sum_{i=1}^{n} x_{i}^{2} \leqslant c n$.
The minimum value of $c$ is $\frac{4}{3}$.
On one hand, given that $x_{1}, x_{2}, \cdots, x_{n}$ are real numbers not less than -1, for any $1 \leqslant i \leqslant n$, we have
$$
\begin{array}{l}
\left(x_{i}+1\right)\left(x_{i}-2\right)^{2} \geqslant 0 \\
\Rightarrow \sum_{i=1}^{n}\left(x_{i}+1\right)\left(x_{i}-2\right)^{2} \geqslant 0 .
\end{array}
$$

Expanding and rearranging the above inequality and noting that $\sum_{i=1}^{n} x_{i}^{3}=0$, we get
$$
\sum_{i=1}^{n} x_{i}^{2} \leqslant \frac{4}{3} n \text {. }
$$

Thus, $c \geqslant \frac{4}{3}$.
On the other hand, when $n=9$, taking $x_{1}=x_{2}=\cdots=x_{8}=-1, x_{9}=2$, we have
$$
\sum_{i=1}^{9} x_{i}^{3}=0 \text {, and } \sum_{i=1}^{9} x_{i}^{2}=12=\frac{4}{3} \times 9,
$$

which means $c=\frac{4}{3}$ satisfies the condition.
In conclusion, the minimum value of $c$ is $\frac{4}{3}$.