Question 2. Given non-negative real numbers $a_{1} \geq a_{2} \geq \ldots \geq a_{n}, b_{1} \leq b_{2} \leq \ldots \leq b_{n}$, satisfying
$$
a_{1} a_{n}+a_{2} a_{n-1}+\ldots+a_{n} a_{1}=b_{1} b_{n}+b_{2} b_{n-1}+\ldots+b_{n} b_{1}=1 \text {, }
$$

Find
$$
S=\sum_{1 \leq i<j \leq n} a_{i} b_{j}
$$

the minimum value.
$$
\begin{aligned}
S & =a_{1}\left(b_{2}+\ldots+b_{n}\right)+a_{2}\left(b_{3}+\ldots+b_{n}\right)+\ldots+a_{n-1} b_{n} \\
& \geq \frac{b_{1}+\ldots+b_{n}}{n} \cdot\left((n-1) a_{1}+(n-2) a_{2}+\ldots+a_{n-1}\right) \\
& \geq \frac{b_{1}+\ldots+b_{n}}{n} \cdot \frac{a_{1}+\ldots+a_{n}}{n} \cdot(1+2+\ldots+(n-1))
\end{aligned}
$$

Also note that
$$
\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{2} \geq n\left(a_{1} a_{n}+a_{2} a_{n-1}+\ldots+a_{n} a_{1}\right)=n,
$$

so $a_{1}+\ldots+a_{n} \geq \sqrt{n}$, and similarly, $b_{1}+\ldots+b_{n} \geq \sqrt{n}$.
Therefore, $S \geq \frac{n-1}{2}$, with equality when $a_{1}=\ldots=a_{n}=b_{1}=\ldots=b_{n}=\frac{\sqrt{n}}{n}$. Hence, the minimum value is $\frac{n-1}{2}$.