Let $f(X)$ be the average of elements of the finite number set $X$.

First of all, make $n$ different primes $p_1, p_2, \dots, p_n$ which are all bigger than $n$, and we prove that for any different nonempty subsets $A$, $B$ of the set $S_1 = \{ \prod (1/p_j) : 1 \le j \le n \}$, $f(A) \ne f(B)$ always holds.

In fact, we can suppose that $\prod_{i=1}^{n} p_i \in A$ and $\prod_{i=1}^{n} p_i \notin B$ without loss of generality. Every element of $B$ can be divided by $p_1$, so $p_1 \mid n! f(B)$. But $A$ has exactly one element which cannot be divided by $p_1$, so we find that $n! f(A)$ cannot be divided by $p_1$ (note that $p_1 > n$), and therefore $n!f(A) \ne n!f(B)$, which follows $f(A) \ne f(B)$.

Second, let $S_2 = \{ n! x : x \in S_1 \}$. Then $f(A)$ and $f(B)$ are different positive integers when $A, B$ are different nonempty subsets of $S_2$.

In fact, it is easy to see that there exist two sets $A_1, B_1$ which are different nonempty subsets of $S_1$, and $f(A) = n! f(A_1)$, $f(B) = n! f(B_1)$ holds. We get $f(A) \ne f(B)$ from $f(A_1) \ne f(B_1)$, and $f(A), f(B)$ are positive integers from $|A|, |B| \le n$ and their elements are all positive.

Then, let $K$ be the largest element of $S_2$. We prove that for every two distinct subsets $A, B$ of the set $S_3 = \{ K! x + 1 : x \in S_2 \}$, $f(A)$ and $f(B)$ are coprime integers which are both larger than $1$.

In fact, it is easy to see that there exist two sets $A_1, B_1$ which are different nonempty subsets of $S_2$, and $f(A) = K! f(A_1) + 1$, $f(B) = K! f(B_1) + 1$ holds. Obviously, $f(A)$ and $f(B)$ are different integers which are both larger than $1$. If they have common divisors, let $p$ be a prime common divisor of them without loss of generality. Clearly, we have $p \mid (K! \cdot |f(A_1) - f(B_1)|)$. We get $1 \le |f(A_1) - f(B_1)| \le K$ by $0 < f(A_1), f(B_1) \le K$ and $f(A_1) \ne f(B_1)$, so $p \le K$, which follows $p \mid K! f(A_1)$, and then $p \mid 1$ — a contradiction.

Lastly, let $L$ be the largest element of $S_3$. We prove that for every two distinct nonempty subsets $A, B$ of the set $S_4 = \{ L! + x : x \in S_3 \}$, $f(A)$ and $f(B)$ are two composites which share no common divisors.

In fact, it is easy to see that there exist two sets $A_1, B_1$ which are different nonempty subsets of $S_3$, and $f(A) = L! + f(A_1)$, $f(B) = L! + f(B_1)$ holds. Obviously, $f(A)$ and $f(B)$ are different integers which are both larger than $1$. Because of that, $L$ is the largest element of $S_3$, and we have $f(A_1) \mid L!$ and $f(A_1) \mid f(A)$. We find that $f(A)$ is composite by $f(A_1) < f(A)$. For a similar reason, $f(B)$ is composite too. If they have common divisors, let $p$ be a prime common divisor of them without loss of generality. It is obvious that $p \mid (L! \cdot |f(A_1) - f(B_1)|)$. We get $1 \le |f(A_1) - f(B_1)| \le L$ by $0 < f(A_1), f(B_1) \le L$ and $f(A_1) \ne f(B_1)$, so $p \le L$, which follows $p \mid f(A_1)$ and $p \mid f(B_1)$ — a contradiction of the fact that $f(A_1)$ and $f(B_1)$ are coprime. This completes the proof.