2. 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 $a_{1}+a_{2}+\cdots+a_{2 n}=b_{1}+b_{2}+\cdots+b_{2 n}>0$, and for any $i=1,2, \cdots, 2 n$, $a_{i} a_{i+2} \geqslant b_{i}+b_{i+1}$ (here $a_{2 n+1}=a_{1}, a_{2 n+2}=a_{2}, b_{2 n+1}=b_{1}$ ).
Find the minimum value of $a_{1}+a_{2}+\cdots+a_{2 n}$.
Let $S=\sum_{k=1}^{2 n} a_{k}=\sum_{k=1}^{2 n} b_{k}, T=\sum_{k=1}^{n} a_{2 k-1}$, without loss of generality, assume $T \leqslant \frac{S}{2}$.
When $n=3$, since $\left(a_{1}+a_{3}+a_{5}\right)^{2}-3\left(a_{1} a_{3}+a_{3} a_{5}+a_{5} a_{1}\right)$
$=a_{1}^{2}+a_{3}^{2}+a_{5}^{2}-\left(a_{1} a_{3}+a_{3} a_{5}+a_{5} a_{1}\right)=\frac{1}{2}\left[\left(a_{1}-a_{3}\right)^{2}+\left(a_{3}-a_{5}\right)^{2}+\left(a_{5}-a_{1}\right)^{2}\right] \geqslant 0$,
thus $\frac{S^{2}}{4} \geqslant T^{2} \geqslant 3\left(a_{1} a_{3}+a_{3} a_{5}+a_{5} a_{1}\right) \geqslant 3\left(b_{1}+b_{2}+b_{3}+b_{4}+b_{5}+b_{6}\right)=3 S$
$\Rightarrow S \geqslant 12$, equality holds when $a_{k}=b_{k}=2, k=1,2, \cdots, 6$.
When $n \geqslant 4$, if $n$ is even, then
$$
\begin{array}{l}
\sum_{k=1}^{n} a_{2 k-1} a_{2 k+1} \leqslant\left(a_{1}+a_{5}+\cdots+a_{2 n-3}\right)\left(a_{3}+a_{7}+\cdots+a_{2 n-1}\right) \\
\leqslant\left(\frac{a_{1}+a_{3}+a_{5}+a_{7}+\cdots+a_{2 n-3}+a_{2 n-1}}{2}\right)^{2}=\frac{T^{2}}{4},
\end{array}
$$

if $n$ is odd, without loss of generality, assume $a_{1} \leqslant a_{3}$, then $\sum_{k=1}^{n} a_{2 k-1} a_{2 k+1} \leqslant \sum_{k=1}^{n-1} a_{2 k-1} a_{2 k+1}+a_{2 n-1} a_{3}$
$$
\leqslant\left(a_{1}+a_{5}+\cdots+a_{2 n-1}\right)\left(a_{3}+a_{7}+\cdots+a_{2 n-3}\right) \leqslant \frac{T^{2}}{4} \text {. }
$$

Thus $\frac{S^{2}}{16} \geqslant \frac{T^{2}}{4} \geqslant \sum_{k=1}^{n} a_{2 k-1} a_{2 k+1} \geqslant \sum_{k=1}^{2 n} b_{k}=S \Rightarrow S \geqslant 16$, equality holds when, in summary, when $n=3$, the minimum value of $S$ is 12; when $n \geqslant 4$, the minimum value of $S$ is 16.