Let $n \geqslant 2$ be a positive integer, and let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers and $b_{1}, b_{2}, \cdots, b_{n}$ be non-negative real numbers satisfying
(a) $a_{1}+a_{2}+\cdots+a_{n}+b_{1}+b_{2}+\cdots+b_{n}=n$;
(b) $a_{1} a_{2} \cdots a_{n}+b_{1} b_{2} \cdots b_{n}=\frac{1}{2}$.

Find the maximum value of $a_{1} a_{2} \cdots a_{n}\left(\frac{b_{1}}{a_{1}}+\frac{b_{2}}{a_{2}}+\cdots+\frac{b_{n}}{a_{n}}\right)$.
(Tian Kaibin, Chu Xiaoguang, Pan Chenghua)
According to the AM-GM inequality, we have
$$
\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right) \cdots\left(a_{n}+b_{n}\right) \leqslant\left(\frac{\left(a_{1}+b_{1}\right)+\left(a_{2}+b_{2}\right)+\cdots+\left(a_{n}+b_{n}\right)}{n}\right)^{n}=1 .
$$

Thus,
$$
a_{1} a_{2} \cdots a_{n}+b_{1} b_{2} \cdots b_{n}+\left(b_{1} a_{2} a_{3} \cdots a_{n}+a_{1} b_{2} a_{3} \cdots a_{n}+\cdots+a_{1} a_{2} a_{3} \cdots b_{n}\right) \leqslant 1 .
$$

That is,
$$
a_{1} a_{2} \cdots a_{n}\left(\frac{b_{1}}{a_{1}}+\frac{b_{2}}{a_{2}}+\cdots+\frac{b_{n}}{a_{n}}\right) \leqslant 1-\left(a_{1} a_{2} \cdots a_{n}+b_{1} b_{2} \cdots b_{n}\right)=\frac{1}{2} .
$$

Taking $a_{1}=a_{2}=\cdots=a_{n-1}=1, a_{n}=\frac{1}{2}, b_{1}=b_{2}=\cdots=b_{n-1}=0, b_{n}=\frac{1}{2}$, satisfies the given conditions.
And it makes $a_{1} a_{2} \cdots a_{n}\left(\frac{b_{1}}{a_{1}}+\frac{b_{2}}{a_{2}}+\cdots+\frac{b_{n}}{a_{n}}\right)=\frac{1}{2}$.
Therefore, the maximum value of $a_{1} a_{2} \cdots a_{n}\left(\frac{b_{1}}{a_{1}}+\frac{b_{2}}{a_{2}}+\cdots+\frac{b_{n}}{a_{n}}\right)$ is $\frac{1}{2}$.