2. Let $n \geqslant 3, \omega=\cos \frac{2 \pi}{n}+\mathrm{i} \sin \frac{2 \pi}{n}$ be an $n$-th root of unity, and let $x_{i}(i=0,1, \cdots, n-1)$ be real numbers, with $x_{0} \geqslant x_{1} \geqslant \cdots \geqslant$ $x_{n-1}$. Find the necessary and sufficient conditions that $x_{0}, x_{1}, \cdots, x_{n-1}$ must satisfy for the following equation to hold: $x_{0}+x_{1} \omega+x_{2} \omega^{2}+\cdots+x_{n-1} \omega^{n-1}=0$.
2. From the problem, we have $\sum_{i=1}^{n-1} x_{i} \sin \frac{2 i \pi}{n}=0$.
(i) When $n$ is even, i.e., $n=2 k, k \in \mathbf{N}_{+}$, we have
$$
\begin{aligned}
& \left(x_{1}-x_{2 k-1}\right) \sin \frac{2 \pi}{2 k}+\left(x_{2}+x_{2 k-2}\right) \sin \frac{2 \cdot 2 \pi}{2 k}+\cdots+\left(x_{k-1}-x_{k+1}\right) \sin \frac{2 \cdot(k-1) \pi}{2 k}+x_{k} \sin \frac{2 k \pi}{2 k} \\
= & \left(x_{1}-x_{2 k-1}\right) \sin \frac{2 \pi}{2 k}+\left(x_{2}-\dot{x}_{2 k-2}\right) \sin \frac{2 \cdot 2 \pi}{2 k}+\cdots+\left(x_{k-1}-x_{k+1}\right) \sin \frac{2(k-1) \pi}{2 k}=0 . \\
\text { Also, } & x_{1} \geqslant x_{2 k-1}, x_{2} \geqslant x_{2 k-2}, \cdots, x_{k-1} \geqslant x_{k+1}, \sin \frac{2 \pi}{2 k}>0, \sin \frac{2 \cdot 2 \pi}{2 k}>0, \cdots, \sin \frac{2(k-1) \pi}{2 k}>0, \\
\therefore \quad & x_{1}=x_{2 k-1}, x_{2}=x_{2 k-2}, \cdots, x_{k-1}=x_{k+1} . \\
\text { Also, } & x_{2 k-1} \leqslant x_{2 k-2} \leqslant \cdots \leqslant x_{1} \leqslant x_{0}, \\
\therefore \quad & x_{1}=x_{2}=\cdots=x_{2 k-1} .
\end{aligned}
$$

Furthermore, from $x_{1}+x_{1} \omega+x_{2} \omega^{2}+\cdots+x_{n-1} \omega^{n-1}=0 \Rightarrow x_{0}+x_{1}\left(\omega+\omega^{2}+\cdots+\omega^{n-1}\right)=0 \Rightarrow x_{0}=x_{1}$.
$$
\therefore x_{0}=x_{1}=\cdots=x_{n-1} \text {. }
$$

When $x_{0}=x_{1}=\cdots=x_{n-1}$, it is clear that $x_{0}+x_{1} \omega+x_{2} \omega^{2}+\cdots+x_{n-1} \omega^{n-1}=0$.
(II) When $n$ is odd, i.e., $n=2 k+1$, we have
$$
\left(x_{1}-x_{2 k}\right) \sin \frac{2 \pi}{2 k+1}+\left(x_{2}-x_{2 k-1}\right) \sin \frac{2-2 \pi}{2 k+1}+\cdots+\left(x_{k}-x_{k+1}\right) \sin \frac{2 \pi}{2 k+1}=0 \text {. }
$$

Similarly to (i), we have $x_{1}=x_{2}=\cdots=x_{2 k+1}$, and $x_{0}=x_{1}$.
$\therefore x_{0}=x_{1}=\cdots=x_{2 k+1}$, and when $x_{0}=x_{1}=\cdots=x_{2 k+1}$, it is clear that $x_{0}+x_{1} \omega+\cdots+x_{n-1} \omega^{n-1}=0$.
In summary, the necessary and sufficient condition for $x_{0}, x_{1}, x_{2}, \cdots, x_{n-1}$ is $x_{0}=x_{1}=x_{2}=\cdots=x_{n-1}$.