Given an integer $n \geqslant 3$. Let $a_{1}, a_{2}, \cdots, a_{2 n}, b_{1}, b_{2}, \cdots, b_{2 n}$ be $4 n$ non-negative real numbers, satisfying
$$
a_{1}+a_{2}+\cdots+a_{2 n}=b_{1}+b_{2}+\cdots+b_{2 n}>0 \text {, }
$$
and for any $i=1,2, \cdots, 2 n$, we have $a_{i} a_{i+2} \geqslant b_{i}+b_{i+1}$, where $a_{2 n+1}=a_{1}, a_{2 n+2}=a_{2}, b_{2 n+1}=b_{1}$. Find the minimum value of $a_{1}+a_{2}+\cdots+a_{2 n}$.
Let $S=a_{1}+a_{2}+\cdots+a_{2 n}=b_{1}+b_{2}+\cdots+b_{2 n}$. Without loss of generality, assume
$$
T=a_{1}+a_{3}+\cdots+a_{2 n-1} \leqslant \frac{S}{2} \text {. }
$$

When $n=3$, because
$$
\begin{aligned}
T^{2} & -3 \sum_{k=1}^{3} a_{2 k-1} a_{2 k+1} \\
& =\frac{1}{2}\left(a_{1}-a_{3}\right)^{2}+\left(a_{3}-a_{5}\right)^{2}+\left(a_{5}-a_{1}\right)^{2} \\
& \geqslant 0,
\end{aligned}
$$

thus, combining the conditions, we have
$$
\begin{array}{l}
\frac{S^{2}}{4} \geqslant T^{2} \geqslant 3 \sum_{k=1}^{3} a_{2 k-1} a_{2 k+1} \\
\geqslant 3 \sum_{k=1}^{3}\left(b_{2 k-1}+b_{2 k}\right)=3 S .
\end{array}
$$

Since $S>0$, we have $S \geqslant 12$.
When $a_{i}=b_{i}=2(1 \leqslant i \leqslant 6)$, $S$ achieves the minimum value of 12.
When $n \geqslant 4$, on one hand,
$$
\sum_{k=1}^{n} a_{2 k-1} a_{2 k+1} \geqslant \sum_{k=1}^{n}\left(b_{2 k-1}+b_{2 k}\right)=S \text {. }
$$

On the other hand, if $n$ is even, then
$$
\begin{array}{l}
\sum_{k=1}^{n} a_{2 k-1} a_{2 k+1} \\
\leqslant\left(a_{1}+a_{5}+\cdots+a_{2 n-3}\right)\left(a_{3}+a_{7}+\cdots+a_{2 n-1}\right) \\
\leqslant \frac{T^{2}}{4},
\end{array}
$$

where the first inequality is because
$$
\left(a_{1}+a_{5}+\cdots+a_{2 n-3}\right)\left(a_{3}+a_{7}+\cdots+a_{2 n-1}\right)
$$

expands to non-negative terms, including $a_{2 k-1} a_{2 k+1}$ $(1 \leqslant k \leqslant n)$, and the second inequality uses the basic inequality;
if $n$ is odd, assume $a_{1} \leqslant a_{3}$, then
$$
\begin{array}{l}
\sum_{k=1}^{n} a_{2 k-1} a_{2 k+1} \\
\leqslant\left(\sum_{k=1}^{n-1} a_{2 k-1} a_{2 k+1}\right)+a_{2 n-1} a_{3} \\
\leqslant\left(a_{1}+a_{5}+\cdots+a_{2 n-3}\right)\left(a_{3}+a_{7}+\cdots+a_{2 n-3}\right) \\
\leqslant \frac{T^{2}}{4} .
\end{array}
$$

Thus, $S \leqslant \sum_{k=1}^{n} a_{2 k-1} a_{2 k+1} \leqslant \frac{T^{2}}{4} \leqslant \frac{S^{2}}{16}$.
Since $S>0$, we have $S \geqslant 16$.
When $a_{1}=a_{2}=a_{3}=a_{4}=4, a_{i}=0(5 \leqslant i \leqslant 2 n), b_{1}=0, b_{2}=16, b_{i}=0(3 \leqslant i \leqslant 2 n)$, $S$ achieves the minimum value of 16.

In summary, when $n=3$, the minimum value of $S$ is 12; when $n \geqslant 4$, the minimum value of $S$ is 16.