China Hong Kong Mathematical Olympiad

Firstly, note that $2005 \equiv 1 \pmod{3}$. Therefore, $3 \mid 2005 - 1$. Suppose we have chosen distinct primes $p_1, p_2, \dots, p_k$ such that $p_1 p_2 \cdots p_k \mid 2005^{p_1 p_2 \cdots p_{k-1}} - 1$, where the exponent of $2005$ is $1$ when $k=1$. Then
$$2005^{p_1 p_2 \cdots p_k} = (2005^{p_1 p_2 \cdots p_{k-1}})^{p_k} \equiv 1^{p_k} = 1 \pmod{p_1 p_2 \cdots p_k}.$$
This shows $n = p_1 p_2 \cdots p_k$ satisfies the requirement.
By Zsigmondy's theorem, there exists a prime $p_k$ dividing $2005^{p_1 p_2 \cdots p_k} - 1$ but not $2005^{p_1 p_2 \cdots p_{k-1}} - 1$. This shows $p_k$ is different from $p_1, p_2, \dots, p_{k-1}$. Therefore, we have
$$p_1 p_2 \cdots p_{k+1} \mid 2005^{p_1 p_2 \cdots p_k} - 1.$$

By induction, we can find an infinite sequence of distinct primes $p_j$ such that $n = p_1 p_2 \cdots p_k$ satisfies the requirement for any $k \in \mathbb{Z}^+$. So we are done.