Example 3 If a positive integer has eight positive divisors, and the sum of these eight positive divisors is 3240, then this positive integer is called a "good number". For example, 2006 is a good number, because the sum of its divisors 1, $2,17,34,59,118,1003,2006$ is 3240. Find the smallest good number. ${ }^{[3]}$
(2006, Brazilian Mathematical Olympiad)
\left(1+\alpha_{1}\right)\left(1+\alpha_{2}\right) \cdots\left(1+\alpha_{k}\right)=8.
Therefore, when $k=1$, $\alpha_{1}=7$;
when $k=2$, $\alpha_{1}=1, \alpha_{2}=3$ or $\alpha_{1}=3, \alpha_{2}=1$;
when $k=3$, $\alpha_{1}=\alpha_{2}=\alpha_{3}=1$;
when $k \geqslant 4$, there is no solution.
(1) If $n=p^{7}$ ( $p$ is a prime number), then $\sum_{i=0}^{7} p^{i}=3240$.
(i) If $p \geqslant 3$, then
$$
\sum_{j=0}^{7} p^{j} \geqslant \sum_{j=0}^{7} y^{j}=\frac{3^{8}-1}{2}=3280>3240 \text {. }
$$
(ii) If $p=2$, then
$$
\sum_{j=0}^{7} p^{j}=2^{8}-1=511 \neq 3240 \text {. }
$$

Therefore, when $n=p^{7}$, there is no solution that satisfies the problem.
(2) If $n=p^{3} q(p, q$ are prime numbers, and $p \neq q)$, then the sum of positive divisors is
$$
\begin{array}{l}
3240=\left(1+p+p^{2}+p^{3}\right)(1+q) \\
=(1+p)\left(1+p^{2}\right)(1+q) \\
=2^{3} \times 3^{4} \times 5 .
\end{array}
$$

Since $q \geqslant 2$, we have $1+q \geqslant 3$.
Thus, $1+p+p^{2}+p^{3} \leqslant 1080$. Therefore, $p \leqslant 7$.
(i) If $p=7$, then $1+p^{2}=50 \times 3240$, which contradicts equation (1).
(ii) If $p=5$, then $1+p^{2}=26 \times 3240$, which is a contradiction.
(iii) If $p=3$, then $1+q=\frac{3240}{40}=81$, thus, $q=80$, which is a contradiction.
(iv) If $p=2$, then $1+q=\frac{3240}{15}=216$, thus, $q=215$, which is also a contradiction.
(3) If $n=p q r(p<q<r)$, then
$(1+p)(1+q)(1+r)$
$=3240=2^{3} \times 3^{4} \times 5$.
(i) If $p=2$, then
$(1+q)(1+r)=2^{3} \times 3^{3} \times 5$.
To make $p q r$ as small as possible, $q r$ should be as small as possible. Thus, from
$q r=(1+q)(1+r)-(q+1)-(r+1)+1$, we know that $(q+1)+(r+1)$ should be as large as possible.

Considering $(1+q)(1+r)=2^{3} \times 3^{3} \times 5$, we know that $q+1$ should be as small as possible, and $r+1$ should be as large as possible.
Therefore, take $q=3, r=269$.
At this point, $n_{\text {min }}=2 \times 3 \times 269=1614$.
(ii) If $p \geqslant 3$, then
$\frac{1+p}{2} \cdot \frac{1+q}{2} \cdot \frac{1+r}{2}=3^{4} \times 5$.
Due to the size relationship of $p, q, r$, there are only two cases:
$\frac{1+p}{2}=3, \frac{1+q}{2}=3^{2}, \frac{1+r}{2}=3 \times 5$;

and $\frac{1+p}{2}=3, \frac{1+q}{2}=5, \frac{1+r}{2}=3^{3}$.
It is easy to solve the corresponding
$(p, q, r)=(5,17,29)$ or $(5,9,53)$.
Since 9 is not a prime number, $(p, q, r)=(5,17,29)$.
Therefore, $n=2465$.
In summary, the $n_{\min }=1614$ that satisfies the problem.