Obviously, the set $\{100, 101, 102, \ldots, 10000\}$ with 9901 elements works, because any two elements have a product more than $10100$. To prove that 9901 is the maximum: Firstly, if $1$ is in $X$, then it is the only element. So $1$ can't be used. Next, consider the three numbers $(i, 199-i, i(199-i))$ for $i = 2, 3, \ldots, 99$. The three numbers can't all be in $X$, or the property would be violated. Furthermore the $98$ triples are disjoint, so at least one number from each triple must be eliminated. There are $98$ triples so $98$ numbers are eliminated, plus $1$ from above, so at least $99$ numbers must be unused.