1. Given an integer $n(n \geqslant 3)$. Find the largest real number $M$, such that for any sequence of positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$, there exists a permutation $y_{1}, y_{2}, \cdots, y_{n}$, satisfying
$$
\sum_{i=1}^{n} \frac{y_{i}^{2}}{y_{i+1}^{2}-y_{i+1} y_{i+2}+y_{i+2}^{2}} \geqslant M,
$$

where, $y_{n+1}=y_{1}, y_{n+2}=y_{2}$.
1. Let
$$
\begin{array}{l}
F\left(x_{1}, x_{2}, \cdots, x_{n}\right) \\
=\sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i+1}^{2}-x_{i+1} x_{i+2}+x_{i+2}^{2}} .
\end{array}
$$

First, take $x_{1}=x_{2}=\cdots=x_{n-1}=1, x_{n}=\varepsilon$. In this case, all permutations are the same under cyclic permutation, then
$$
\begin{array}{l}
F\left(x_{1}, x_{2}, \cdots, x_{n}\right) \\
=n-3+\frac{2}{1-\varepsilon+\varepsilon^{2}}+\varepsilon^{2} .
\end{array}
$$

Let $\varepsilon \rightarrow 0^{+}$. Hence, the above expression $\rightarrow n-1$.
Thus, $M \leqslant n-1$.
Next, we prove: For any positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$, there exists a permutation $y_{1}, y_{2}, \cdots, y_{n}$, such that
$$
F\left(y_{1}, y_{2}, \cdots, y_{n}\right) \geqslant n-1 \text {. }
$$

In fact, take the permutation $y_{1}, y_{2}, \cdots, y_{n}$ satisfying
$$
y_{1} \geqslant y_{2} \geqslant \cdots \geqslant y_{n} \text {. }
$$

Using the inequality
$$
a^{2}-a b+b^{2} \leqslant \max \left\{a^{2}, b^{2}\right\}
$$

for positive real numbers $a, b$, we know
$$
\begin{array}{l}
F\left(y_{1}, y_{2}, \cdots, y_{n}\right) \geqslant \frac{y_{1}^{2}}{y_{2}^{2}}+\frac{y_{2}^{2}}{y_{3}^{2}}+\cdots+\frac{y_{n-1}^{2}}{y_{1}^{2}} \\
\geqslant n-1 .
\end{array}
$$

The last inequality is the AM-GM inequality.
In summary, $M=n-1$.