3. Given an integer $n \geqslant 2$. Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}>0$, satisfying
$$
a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n} \text {, }
$$

and for any $i, j(1 \leqslant i<j \leqslant n)$, we have $a_{i} a_{j} \geqslant b_{i}+b_{j}$. Find the minimum value of $a_{1}+a_{2}+\cdots+a_{n}$.
$\sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}$ consists of $C_{n}^{2}=\frac{n(n-1)}{2}$ terms, where each $a_{i}$ appears $n-1$ times, then we have $\sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j} \geqslant(n-1) \sum_{i=1}^{n} b_{i}=(n-1) \sum_{i=1}^{n} a_{i}$.
On the other hand, $\sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j} \leqslant \frac{1}{2} \sum_{1 \leqslant i<j \leqslant n}\left(a_{i}^{2}+a_{j}^{2}\right)=\frac{n-1}{2} \sum_{i=1}^{n} a_{i}^{2}$,
thus $\left(\sum_{i=1}^{n} a_{i}\right)^{2}=\sum_{i=1}^{n} a_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j} \geqslant\left(\frac{2}{n-1}+2\right) \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}$ $=\frac{2 n}{n-1} \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j} \geqslant 2 n \sum_{i=1}^{n} a_{i} \Rightarrow \sum_{i=1}^{n} a_{i} \geqslant 2 n$,
equality holds when $a_{i}=b_{i}=2, i=1,2, \cdots, n$. Therefore, the minimum value of $\sum_{i=1}^{n} a_{i}$ is $2 n$.