4.3.4.4 * A natural number greater than 1, if it is exactly equal to the product of its distinct proper divisors (divisors other than 1 and itself), then it is called "good". Find the sum of the first ten "good" natural numbers.
Let $n$ be a natural number, and its distinct proper divisors are arranged in increasing order as $d_{1}, d_{2}, \cdots, d_{k}$. Then, it must be true that $1 \cdot n=d_{1} \cdot d_{k}=d_{2} \cdot d_{k-1}=\cdots$. If $n$ is "good," then it should also be true that $1 \cdot n=d_{1} \cdot d_{2} \cdots d_{k}$. Therefore, $k$ can only be $2$, and $d_{1}$ must be a prime number; $d_{2}$ must be a prime number or $d_{1}^{2}$. That is, $n$ can only be the product of two primes or the cube of a prime. Thus, the first ten "good" natural numbers can be found: $6,8,10,14,15,21$, $22,26,27,33$, and their sum is 182.