Example 1 Let the positive integer $n$ be a multiple of 75, and have exactly 75 positive divisors (including 1 and itself). Find the minimum value of $n$.
Solution: Let the prime factorization of $n$ be
$$
n=p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{k}^{r_{k}},
$$

where $p_{1}, p_{2}, \cdots, p_{k}$ are the distinct prime factors of $n$, and $r_{1}, r_{2}, \cdots, r_{k}$ are all positive integers.
Thus, the number of distinct positive divisors of $n$ is
$$
\left(r_{1}+1\right)\left(r_{2}+1\right) \cdots\left(r_{k}+1\right).
$$

From the problem, we have
$$
\left(r_{1}+1\right)\left(r_{2}+1\right) \cdots\left(r_{k}+1\right)=75=3 \times 5^{2}.
$$

Therefore, $n$ can have at most 3 distinct prime factors.
To make $n$ the smallest and a multiple of 75, the prime factors of $n$ should be $2, 3, 5$, and 3 should appear at least once, and 5 should appear at least twice, i.e.,
$$
\begin{array}{l}
n=2^{r_{1}} \times 3^{r_{2}} \times 5^{r_{3}}, \\
\left(r_{1}+1\right)\left(r_{2}+1\right)\left(r_{3}+1\right)=75, \\
r_{1} \geqslant 0, r_{2} \geqslant 1, r_{3} \geqslant 2 .
\end{array}
$$

The tuples $\left(r_{1}, r_{2}, r_{3}\right)$ that satisfy the above conditions are
$$
\begin{array}{l}
(4,4,2),(4,2,4),(2,4,4),(0,4,14), \\
(0,14,4),(0,2,24),(0,24,2) .
\end{array}
$$

After calculation, it is found that when $r_{1}=r_{2}=4, r_{3}=2$, $n$ takes the minimum value, which is $n=2^{4} \times 3^{4} \times 5^{2}=32400$.