Solution. It is clear that, if we remove 43 numbers $2,3, \ldots, 44$, then, since $45^{2}=$ 2025, among those left no one is the product of any two others. This is the minimal number. To prove that consider 43 triples $(k, 89-k,(89-k) k)$, for $k=2, \ldots, 44$. They do not have numbers in common and we have to remove at least one number from every such triple.
8. How few numbers is it possible to cross out from the sequence
$$
1,2,3, \ldots, 2023
$$
so that among those left no number is the product of any two (distinct) other numbers?