Example 1 If a positive integer has 21 positive divisors, what is the smallest positive integer with this property?
Analysis We handle this using the Fundamental Theorem of Arithmetic.
Solution This problem is solved in two steps.
First, let the prime factorization of the positive integer $N$ be $N=p_{1}^{t} p_{i}^{t} \cdots p_{k}^{t}$.
Where $l_{1} \geqslant 1(i=1,2, \cdots, k), p_{1}, p_{2}, \cdots, p_{k}$ are distinct primes. We calculate the number of positive divisors of $N$. Clearly, the positive divisors of $N$ are $d=p_{1}^{m_{5}} p_{2}^{m_{2}} \cdots p_{k^{\prime}}^{m^{\prime}}$, where $0 \leqslant m_{1} \leqslant l_{1}, 0 \leqslant m_{2} \leqslant l_{2}, \cdots, 0 \leqslant m_{k} \leqslant l_{k}$, when $m_{i}$ takes all values $0,1,2, \cdots, l_{i}$, $d$ takes all positive divisors of $N$, thus, the number of positive divisors of $N$ is
$$
\prod_{i=1}^{k}\left(l_{i}+1\right)=\left(l_{1}+1\right)\left(l_{2}+1\right) \cdots\left(l_{k}+1\right) .
$$

Second, suppose the positive integer $N$ has 21 positive divisors. If we still set $N=p_{1}^{\prime} p_{i}^{\prime} \cdots p_{k}^{\prime}$. Then we have
$$
\left(l_{1}+1\right)\left(l_{2}+1\right) \cdots\left(l_{k}+1\right)=21=3 \times 7 \times 1 \times \cdots \times 1 .
$$

So $l_{1}+1=3, l_{2}+1=7, l_{3}+1=1, \cdots, l_{k}+1=1$.
We get $l_{1}=2, l_{2}=6, l_{3}=\cdots=l_{k}=0$.
Therefore, a positive integer $N$ with 21 positive divisors must have the form: $N=p_{1}^{2} \cdot p_{2}^{6}$.
If we take $p_{1}=3, p_{2}=2$, then $N=2^{6} \cdot 3^{2}=576$ is the smallest positive integer that satisfies the condition.