HMMT February

First, we construct an example for $N=4$. Let $X_{1}, X_{2}, X_{3}, X_{4}$ be pairwise disjoint sets such that $X_{1}=\varnothing,\left|X_{2}\right|=1,\left|X_{3}\right|=2$, and $\left|X_{4}\right|=2$. It is straightforward to verify the condition. We claim that there are no five sets $X_{1}, X_{2}, \ldots, X_{5}$ for which $\text{(X a X b X c )= a b c}$, for $1 \leq a<b<c \leq 5$. Note that showing the non-existence of five such sets implies that there are no $n$ sets with the desired property for $n \geq 5$ as well. Suppose, for sake of contradiction, that there are such $X_{1}, \ldots, X_{5}$. Then, note that $\left|X_{1} \cup X_{2} \cup X_{4}\right|=3$, $\left|X_{1} \cup X_{2} \cup X_{5}\right|=4$, and $\left|X_{2} \cup X_{4} \cup X_{5}\right|=7$. Note that $\left|X_{1} \cup X_{2} \cup X_{4}\right|+\left|X_{1} \cup X_{2} \cup X_{5}\right|=\left|X_{2} \cup X_{4} \cup X_{5}\right|$. For any sets $A, B, C, D$, we have the following two inequalities: $|A \cup B \cup C|+|A \cup B \cup D| \geq|A \cup B \cup C \cup D| \geq|B \cup C \cup D|$. For $A=X_{1}, B=X_{2}, C=X_{4}$, and $D=X_{5}$ in the situation above, we conclude that the equalities must both hold in both inequalities. The first equality shows that $X_{1} \cup X_{2}=\varnothing$, and therefore both $X_{1}$ and $X_{2}$ are empty. Now observe that $\left|X_{1} \cup X_{4} \cup X_{5}\right|=5 \neq 7=\left|X_{2} \cup X_{4} \cup X_{5}\right|$. This gives a contradiction.