When $n=2$, $\left(a_{1}+a_{2}\right)^{2}-\left(a_{1}-a_{2}\right)^{2}=4 a_{1} a_{2}=4 b_{1} b_{2}=\left(b_{1}+b_{2}\right)^{2}-\left(b_{1}-b_{2}\right)^{2}$, and $a_{1}-a_{2} \leqslant b_{1}-b_{2}$, so $a_{1}+a_{2} \leqslant b_{1}+b_{2}$.

Assume $n \geqslant 3$. Without loss of generality, let $b_{1} b_{2} \cdots b_{n}=1$ (otherwise, use $a_{1}^{\prime}=\frac{a_{1}}{a_{1} a_{2} \cdots a_{n}}, b_{1}^{\prime}=\frac{b_{1}}{b_{1} b_{2} \cdots b_{n}}$ to replace $a, b, 1 \leqslant i \leqslant n$), and $a_{1} \leqslant n-1$, then $\sum_{i=1}^{n} a_{i} \leqslant n(n-1) \leqslant (n-1) \sum_{i=1}^{n} b_{i}$ (this uses the mean inequality). Below, assume $a_{1} > n-1$, because $\sum_{1 \leqslant i \leq n}\left(a_{i}-a_{n}\right) \geqslant \sum_{i=1}^{n} a_{i} + (a_{1}-a_{n-1}) - n a_{n}, \sum_{1 \leqslant i \leq n} (n-1) b_{i} + 2 b_{2} \geqslant 2 n b_{n} \\
a_{n} > 2 b_{n}
\end{array}
Also, from $a_{1} a_{2} \cdots a_{n}=1$ we get $a_{n} \leqslant 1$, so $a_{1} - (n-1) a_{n} > n-1 - (n-1) = 0$, combining with (1) we get $2 b_{2} < 2 a_{1}$. Therefore,
(n-1) \sum_{i=1}^{n} b_{i} > 2(n-1) a_{1} > n a_{1} > \sum_{i=1}^{n} a_{i} .
Note: The coefficient $n-1$ cannot be changed to $n-2$. Take $h$ sufficiently large $(h=n-1), k > h+1$, and let
a_{1}=a_{2}=\cdots=a_{n-1}=h, a_{n}=\frac{1}{h^{n-1}}, b_{1}=k, b_{2}=\cdots=b_{n-1}=1, b_{n}=\frac{1}{k},
$$$$

then all the given conditions are satisfied, and $\sum_{i=1}^{n} a_{i} \therefore (n-1) h > (n-2) \sum_{i=1}^{n} b_{i}$.