Given a positive integer $n$, let real numbers $a_{1}$, $a_{2}, \cdots, a_{n}$ and positive real numbers $b_{1}, b_{2}, \cdots, b_{n}$ satisfy
$$
\begin{array}{l}
\sum_{i=1}^{n} a_{i} \geqslant 8 \text { and } \sum_{i=1}^{n} b_{i} \leqslant 4 . \\
\text { Find the minimum value of } \prod_{i=1}^{n} \mathrm{e}^{\frac{\left(\max \left|0, a_{i}\right|\right)^{2}}{b_{i}}} \text { . }
\end{array}
$$
The minimum value sought is $\mathrm{e}^{16}$, which is achieved if and only if $a_{i}=2 b_{i}$ and $\sum_{i=1}^{n} a_{i}=8, \sum_{i=1}^{n} b_{i}=4$.

In fact,

it is only necessary to find the minimum value of $\sum_{i=1}^{n} \frac{\left(\max \left\{0, a_{i}\right\}\right)^{2}}{b_{i}}$.
By the Cauchy-Schwarz inequality, we have
$$
\sum_{i=1}^{n} \frac{\left(\max \left\{0, a_{i}\right\}\right)^{2}}{b_{i}} \geqslant \frac{\left(\sum_{i=1}^{n} \max \left\{0, a_{i}\right\}\right)^{2}}{\sum_{i=1}^{n} b_{i}} .
$$

By symmetry, without loss of generality, assume
$$
\begin{array}{l}
a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{r} \leqslant 0 \leqslant a_{r+1} \leqslant a_{r+2} \leqslant \cdots \leqslant a_{n} . \\
\text { Let } \sum_{i=1}^{r} a_{i}=-s0 .
\end{array}
$$

Then $t-s \geqslant 8$, i.e., $t \geqslant 8+s$, and
$$
\begin{array}{l}
\sum_{i=1}^{n} \max \left\{0, a_{i}\right\}=\sum_{i=r+1}^{n} a_{i}=t . \\
\text { Hence } \frac{\left(\sum_{i=1}^{n} \max \left\{0, a_{i}\right\}\right)^{2}}{\sum_{i=1}^{n} b_{i}}=\frac{\left(\sum_{i=r+1}^{n} a_{i}\right)^{2}}{\sum_{i=1}^{n} b_{i}} \\
\geqslant \frac{t^{2}}{4} \geqslant \frac{8^{2}}{4}=16,
\end{array}
$$

The equality holds if and only if $a_{i}=2 b_{i}$ and $\sum_{i=1}^{n} a_{i}=8, \sum_{i=1}^{n} b_{i}=4$.
Therefore, the minimum value of $\prod_{i=1}^{n} e^{\frac{(\operatorname{mam}(0.0, i) 1)^{2}}{\sigma_{i}}}$ is $\mathrm{e}^{16}$.