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 positive divisors $1, 2, 17, 34, 59, 118, 1003, 2006$ is 3240. Find the smallest good number.
Let $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, where $\alpha_{i} \in \mathbf{Z}_{+}, p_{i} (i=1,2, \cdots, k)$ are prime numbers, and $p_{1}<p_{2}<\cdots<p_{k}$.
From $\tau(n)=\prod_{i=1}^{k}\left(1+\alpha_{i}\right)=8=2^{3}$, we know $k=1,2,3$.
(1) If $k=1$, then
$\alpha_{1}=7, n=p^{7}$ ($p$ is a prime number), $\sum_{i=0}^{7} p^{i}=3240$. By $\sum_{i=0}^{7} 2^{i}=255<3240<\sum_{i=0}^{7} 3^{i}=3280$,

we know such a prime number $p$ does not exist.
(2) If $k=2$, then let $n=p^{3} q (p, q$ are prime numbers).
Thus $\left(1+p+p^{2}+p^{3}\right)(1+q)$
$$
=3240=2^{3} \times 3^{4} \times 5.
$$

Since $1+q \geqslant 3$, we have
$$
1+p+p^{2}+p^{3} \leqslant 1080 \Rightarrow p \leqslant 7.
$$
(i) If $p=2$, then $1+q=216 \Rightarrow q=215$;
(ii) If $p=3$, then $1+q=81 \Rightarrow q=80$;
(iii) If $p=5$, then $1+q=\frac{270}{13} \notin \mathbf{Z}$;
(iv) If $p=7$, then $1+q=\frac{41}{5} \notin \mathbf{Z}$.

All these cases contradict the fact that $q$ is a prime number.
(3) If $k=3$, then let
$n=p q r (p, q, r$ are prime numbers, $p<q<r)$.
Thus $(1+p)(1+q)(1+r)=3240$.
(i) If $p=2$, then
$$
(1+q)(1+r)=1080=2^{3} \times 3^{3} \times 5 \text{ (it is easy to see that }
$$

$1+q, 1+r$ are both even numbers)
$$
\begin{aligned}
\Rightarrow & (1+q, 1+r) \\
= & (4,270),(6,180),(10,108),(12,90), \\
& (18,60),(20,54),(30,36) \\
\Rightarrow & (q, r)=(3,269),(5,179),(11,89), \\
& (17,59),(19,53) \\
\Rightarrow & n=2 \times 3 \times 269=1614 \text{ or } \\
n= & 2 \times 5 \times 179=1790 \text{ or } \\
n= & 2 \times 11 \times 89=1958 \text{ or } \\
n= & 2 \times 17 \times 59=2006 \text{ or } \\
n= & 2 \times 19 \times 53=2014 .
\end{aligned}
$$
(ii) If $p \geqslant 3$, then $1+p, 1+q, 1+r$ are all even numbers.

Let $1+p=2 p_{1}, 1+q=2 q_{1}, 1+r=2 r_{1}$.
Then $p_{1} q_{1} r_{1}=405=3^{4} \times 5 (2 \leqslant p_{1}<q_{1}<r_{1})$ hence $\left(p_{1}, q_{1}, r_{1}\right)=(3,5,27),(3,9,15)$ $\Rightarrow(p, q, r)=(5,9,53),(5,17,29)$.
Since $q$ is a prime number, we get
$$
\begin{array}{l}
(p, q, r)=(5,17,29) \\
\Rightarrow n=5 \times 17 \times 29=2465 .
\end{array}
$$

In summary, $n_{\min }=2 \times 3 \times 269=1614$.