HMMT February, 2018

Solution:
First, we construct an example for $N=4$. Let $X_{1}, X_{2}, X_{3}, X_{4}$ be pairwise disjoint sets such that $X_{1}=\varnothing$, $|X_{2}|=1$, $|X_{3}|=2$, and $|X_{4}|=2$. It is straightforward to verify the condition.

We claim that there are no five sets $X_{1}, X_{2}, \ldots, X_{5}$ for which $\#(X_{a} \cup X_{b} \cup X_{c}) = \lceil \sqrt{a b c} \rceil$, 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 $|X_{1} \cup X_{2} \cup X_{4}| = 3$, $|X_{1} \cup X_{2} \cup X_{5}| = 4$, and $|X_{2} \cup X_{4} \cup X_{5}| = 7$. Note that
$$|X_{1} \cup X_{2} \cup X_{4}| + |X_{1} \cup X_{2} \cup X_{5}| = |X_{2} \cup X_{4} \cup X_{5}|.$$
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 $|X_{1} \cup X_{4} \cup X_{5}| = 5 \neq 7 = |X_{2} \cup X_{4} \cup X_{5}|$. This gives a contradiction.