Proof:
A basic idea is: to find $n$ consecutive positive integers, each of which has two distinct prime factors. For this, for any $n \in \mathbf{N}^{*}$, take $2n$ different primes $p_{1}, p_{2}, \cdots, p_{n}; q_{1}, q_{2}, \cdots, q_{n}$. By the Chinese Remainder Theorem, there exists $m \in \mathbf{N}^{*}$, such that
$$m \equiv -k \left(\bmod p_{k} q_{k}\right), k=1,2, \cdots, n$$

hold simultaneously. Then in the $n$ consecutive positive integers: $m+1, m+2, \cdots, m+n$, each number has two distinct prime factors, and the proposition is proved.