31st Turkish Mathematical Olympiad

Answer: Yes. Let $n$ be a positive integer whose value will be determined later on. Let $S$ be the set consisting of $2^n$'s for $i=0,1,\dots ,22$. Let $Q$ be the set of subsets of $S$ whose product of own elements is a rational number together with the empty set. Since every positive integer has a unique representation as sum of distinct powers of $2$, the cardinality of $Q$ is precisely the number of integers from $0$ to $K=2^{23}-1$ which are divisible by $n$. Therefore, $$|Q|=2423⟺1+\lfloor \frac{K}{n}\rfloor =2423⟺2422\le \frac{K}{n}<2423⟺\frac{K}{2423}<n\le \frac{K}{2422}.$$ Thus, if $K/2422-K/2423\ge 1$, then there exists an integer $n$ such that $|Q|=2423$. As $2422\cdot 2423<(2^{11}\sqrt{2}-1{)}^2<2^{23}-1=K$, there is such an $n$. (Indeed, $n=3463$ fulfills the condition.)