A number theory problem by A Former Brilliant Member

Each number is of the form $2^a 3^b 5^c 7^d$ . If we look at the exponents $(a,b,c,d)$ mod 3, there are $3^4 = 81$ possibilities. So by the pigeonhole principle, there is at least one 4-tuple $(a,b,c,d) \in \{0,1,2\}^4$ with three of the 175 numbers (since $175 > 81*2$ ). Multiply those three together and you get a perfect cube. Showing that $n=1,2$ is straightforward.