10. Given a positive integer $n(n \geqslant 2)$. It is known that $2 n$ positive real numbers $a_{1}, a_{2}, \cdots, a_{2 n}$, satisfy:
$$
\sum_{k=1}^{n} a_{2 k-1} \cdot \sum_{k=1}^{n} a_{2 k}=\prod_{k=1}^{n} a_{2 k-1}+\prod_{k=1}^{n} a_{2 k} .
$$

Find the minimum value of $S=\sum_{k=1}^{2 n} \frac{a_{k}^{n-1}}{a_{k+1}}$, where $a_{2 n+1}=a_{1}$.
$$
\begin{array}{l}
S=\sum_{k=1}^{2 n} \frac{a_{k}^{n-1}}{a_{k+1}}=\sum_{k=1}^{n} \frac{a_{2 k-1}^{n-1}}{a_{2 k}}+\sum_{k=1}^{n} \frac{a_{2 k}^{n-1}}{a_{2 k+1}} \geqslant n\left[\left(\prod_{k=1}^{n} \frac{a_{2 k-1}^{n-1}}{a_{2 k}}\right)^{\frac{1}{n}}+\left(\prod_{k=1}^{n} \frac{a_{2 k}^{n-1}}{a_{2 k+1}}\right)^{\frac{1}{n}}\right] \\
=\frac{n}{\left(\prod_{k=1}^{2 n} a_{k}\right)^{\frac{1}{n}}}\left(\prod_{k=1}^{n} a_{2 k-1}+\prod_{k=1}^{n} a_{2 k}\right)=\frac{n}{\left(\prod_{k=1}^{2 n} a_{k}\right)^{\frac{1}{n}}}\left(\sum_{k=1}^{n} a_{2 k-1} \cdot \sum_{k=1}^{n} a_{2 k}\right) \\
\geqslant \frac{n}{\left(\prod_{k=1}^{2 n} a_{k}\right)^{\frac{1}{n}}} \cdot\left[n\left(\prod_{k=1}^{n} a_{2 k-1}\right)^{\frac{1}{n}}\right] \cdot\left[n\left(\prod_{k=1}^{n} a_{2 k}\right)^{\frac{1}{n}}\right]=n^{3},
\end{array}
$$

Equality holds when $a_{1}=a_{3}=\cdots=a_{2 n-1}, a_{2}=a_{4}=\cdots=a_{2 n}$, when $n=2$, $2 a_{1} \cdot 2 a_{2}=a_{1}^{2}+a_{2}^{2} \Rightarrow a_{1}^{2}-4 a_{1} a_{2}+a_{2}^{2}=0 \Rightarrow \frac{a_{1}}{a_{2}}=2 \pm \sqrt{3}$, taking $a_{1}=a_{3}=1$, $a_{2}=a_{4}=2+\sqrt{3}$ we get $S=\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\frac{a_{3}}{a_{4}}+\frac{a_{4}}{a_{1}}=2\left(\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{1}}\right)=8 ;$
When $n \geqslant 3$, taking $a_{1}=a_{2}=\cdots=a_{2 n}$, then $n^{2} a_{1}^{2}=2 a_{1}^{n} \Rightarrow a_{1}^{n-2}=\frac{n^{2}}{2}$,
we get $S=\sum_{k=1}^{2 n} a_{1}^{n-2}=2 n \cdot \frac{n^{2}}{2}=n^{3}$, that is, at this time $a_{1}=a_{2}=\cdots=a_{2 n}=\left(\frac{n^{2}}{2}\right)^{\frac{1}{n-2}}$. In summary, the minimum value of $S=\sum_{k=1}^{2 n} \frac{a_{k}^{n-1}}{a_{k+1}}$ is $n^{3}$.