Example 16 (28th Brazilian Mathematical Olympiad) If a positive integer has 8 positive divisors, and the sum of these 8 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.
Let $n=\prod_{i=1}^{k} p_{i}^{a_{i}}$, where $p_{1}3240$.
(ii) If $p=2$, then $\prod_{j=0}^{7} p^{j}=2^{8}-1=511 \neq 3240$.
(2) If $n=p^{3} q, p, q$ are primes, and $p \neq q$, then
$\left(1+p+p^{2}+p^{3}\right)(1+q)=3240$,
$(1+p)\left(1+p^{2}\right)(1+q)=3240=2^{3} \times 3^{4} \times 5$.
Since $q \geqslant 2$, we have $1+q \geqslant 3$, thus,
$1+p+p^{2}+p^{3} \leqslant 1080$.
Therefore, $p^{3} \leqslant 1080 \Rightarrow p \leqslant 10 \Rightarrow p=2,3,5,7$.
(i) If $p=5$, then $1+p^{2}=26 \times 3240$, contradiction.
(ii) If $p=7$, then $1+p^{2}=50 \times 3240$, contradiction.
(iii) If $p=2$, then $1+q=\frac{3240}{3 \times 5}=216$, thus, $q=215$, contradiction.
(iv) If $p=3$, then $1+q=\frac{3240}{4 \times 10}=81$, thus, $q=80$, contradiction.
(3) If $n=p q r$, then
$(1+p)(1+q)(1+r)=3240=2^{3} \times 3^{4} \times 5$.
Assume $p<q<r$.
(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,
$q r=(1+q)(1+r)-(q+1)-(r+1)+1$.
Therefore, $(q+1)+(r+1)$ should be as large as possible.
113
So, $q$ should be as small as possible, and $r$ should be as large as possible.
When $q=3$, $r=269$. At this time,
$$
n_{\min }=2 \times 3 \times 269=1614 \text {. }
$$
(ii) If $p \geqslant 3$, then
$$
\frac{1+p}{2} \cdot \frac{1+q}{2} \cdot \frac{1+r}{2}=3^{4} \times 5 \text {. }
$$

Since $\frac{1+p}{2}, \frac{1+q}{2}, \frac{1+r}{2} \in\left(\mathbf{Z}_{+} \backslash\{1\}\right)$, the possible cases are:
$$
\frac{1+p}{2}=3, \frac{1+q}{2}=5, \frac{1+r}{2}=3^{3} \text { or } \frac{1+p}{2}=3, \frac{1+q}{2}=3^{2}, \frac{1+r}{2}=3 \times 5 \text {. }
$$

The corresponding $(p, q, r)=(5,9,53)$ or $(5,17,29)$.
Since 9 is not a prime, thus,
$$
(p, q, r)=(5,17,29), n=5 \times 17 \times 29=2465 \text {. }
$$

In summary, we know $n_{\min }=1614$.