$$
\begin{array}{l}
\text { Find the largest positive integer } n, \text { such that for positive real } \\
\text { numbers } \alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}, \text { we have } \\
\quad \sum_{i=1}^{n} \frac{\alpha_{i}^{2}-\alpha_{i} \alpha_{i+1}}{\alpha_{i}^{2}+\alpha_{i+1}^{2}} \geqslant 0\left(\alpha_{n+1}=\alpha_{1}\right) .
\end{array}
$$
Let $a_{i}=\frac{\alpha_{i+1}}{\alpha_{i}}(i=1,2, \cdots, n)$.
Then $\prod_{i=1}^{n} a_{i}=1$, and $a_{i}>0$.
The inequality to be proved becomes
$\sum_{i=1}^{n} \frac{1-a_{i}}{1+a_{i}^{2}} \geqslant 0$.
Let $x_{i}=\ln a_{i}(i=1,2, \cdots, n)$.
Then $\sum_{i=1}^{n} \frac{1-\mathrm{e}^{x_{i}}}{1+\left(\mathrm{e}^{x_{i}}\right)^{2}} \geqslant 0$, and $\sum_{i=1}^{n} x_{i}=0$.
Let $f(x)=\frac{1-\mathrm{e}^{x}}{1+\left(\mathrm{e}^{x}\right)^{2}}$.
Differentiation shows that $f(x)$ is decreasing on the interval $(-\infty, \ln (1+\sqrt{2}))$ and increasing on the interval $(\ln (1+\sqrt{2}),+\infty)$.
(1) If there exists $x_{i}x>y$ such that $f^{\prime \prime}(x)>0$.
(2) If there exist $x_{i} 、 x_{j}2$ such that $a_{2}^{8}>2 a_{2}^{7}$.

Hence, inequality (2) $>0$, holds.
(ii) When $0 \leqslant a_{2} \leqslant 2$, $2 a_{2}^{5} \geqslant a_{2}^{6}$.

Hence, inequality (2) $\geqslant a_{2}^{6}\left(a_{2}-1\right)^{2} \geqslant 0$, holds.
(iii) When $a_{2}<0$,
$$
\begin{array}{l}
\text { Inequality (2) }=a_{2}^{8}-2 a_{2}^{7}+2 a_{2}^{5}+a_{2}^{4}+a_{2}^{2}+ \\
2 a_{2}^{2}\left(a_{2}+1\right)^{2}+\left(a_{2}+2\right)^{2} \\
\geqslant a_{2}^{8}-2 a_{2}^{7}+2 a_{2}^{5}+a_{2}^{4}+a_{2}^{2} \\
=a_{2}^{2}\left(a_{2}^{3}+1\right)^{2}-2 a_{2}^{7}+a_{2}^{4} \geqslant 0, \\
\end{array}
$$

also holds.
Therefore, when $n=5$, the original inequality holds.
Thus, the maximum value of $n$ is 5.