Problem 1888

**The answer is $n = 6, 7, 8$.** Note that for $6 \le m \le n$, if we can construct an $\mathcal{F}$ satisfying the problem's conditions for $\{1, 2, \dots, n\}$, then we can necessarily construct an $\mathcal{F}$ satisfying the problem's conditions for $\{1, 2, \dots, m\}$. Therefore, we only need to prove two points:

1. When $n = 9$, there is no $\mathcal{F}$ satisfying the problem's conditions.
Proof: By contradiction, suppose such an $\mathcal{F}$ exists. Then for any $A \subset \{1, 2, \dots, n\}$, there exist three $B \in \mathcal{F}$ satisfying the condition, so the total number of $(A, B)$ combinations is $C_6^n \times 3$. On the other hand, since each $B$ lies in $C_2^{n-2}$ different $A$'s, the total number of $(A, B)$ combinations is also $C_2^{n-2} \times |\mathcal{F}|$.
This means
$$3C_6^n = C_2^{n-2} \times |\mathcal{F}| \Rightarrow |\mathcal{F}| = \frac{1}{5}C_4^n,$$
hence $|\mathcal{F}|$ is not an integer, a contradiction.

2. When $n = 8$, there exists an $\mathcal{F}$ satisfying the problem's conditions.
Let $[n] = \{1, 2, \dots, n\}$, and for any set $X$, let $X_m^n = X \cap \{m, m+1, \dots, n\}$.
Consider the set
$$\mathcal{F} = \{[4], [4]^c\} \cup \{X_1^4 \cup (X_1^4 + 4), X_1^4 \cup ((X^c)_1^4 + 4) : X_1^4 \subset [4], |X_1^4| = 2\}.$$
Now note that if $|A| = 6$ and $A \subset [8]$, then $|A_1^4| = \{2, 3, 4\}$.

- $|A_1^4| = 2$: Without loss of generality, let $A = \{1, 2, 5, 6, 7, 8\}$. In this case, the elements of $\mathcal{F}$ that are subsets of $A$ are exactly $\{5, 6, 7, 8\}$, $\{1, 2, 5, 6\}$, and $\{1, 2, 7, 8\}$, and it is easy to check that this satisfies the problem's conditions.

- $|A_1^4| = 4$: Without loss of generality, let $A = \{1, 2, 3, 4, 7, 8\}$. In this case, the elements of $\mathcal{F}$ that are subsets of $A$ are exactly $\{1, 2, 3, 4\}$, $\{3, 4, 7, 8\}$, and $\{1, 2, 7, 8\}$, and it is easy to check that this satisfies the problem's conditions.

- $|A_1^4| = 3$: In this case $|A_1^4| = |A_5^8| = 3$. Consider two cases:

* $A_5^8 - 4 = A_1^4$: Without loss of generality, let $A = \{1,2,3,5,6,7\}$. In this case, the elements of $\mathcal{F}$ that are subsets of $A$ are exactly $\{1,2,5,6\}$, $\{1,3,5,7\}$, and $\{2,3,6,7\}$, and it is easy to check that this satisfies the problem's conditions.

* $A_5^8 - 4 \neq A_1^4$: Without loss of generality, assume $A = \{1,2,3\}$ and $A_5^8 - 4 = \{1,2,4\}$,
so $A = \{1,2,3,5,6,8\}$. In this case, the elements of $\mathcal{F}$ that are subsets of $A$ are exactly $\{1,2,5,6\}$,
$\{1,3,6,8\}$, and $\{2,3,5,8\}$, and it is easy to check that this satisfies the problem's conditions.