Question 3 Let $n(n>3)$ be a given integer. Find the largest integer $d$, such that for any set $S$ of $n$ integers, there exist four distinct non-empty subsets (which may intersect), the sum of the elements of each of which is divisible by $d$.
Solve for the largest integer $d=n-2$.
Let $S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$.
For convenience, the sum of elements in set $A$ is denoted as $S(A)$.
When $d \geqslant n$, let
$a_{i} \equiv 1(\bmod n)(i=1,2, \cdots, n)$.
Then the sum of elements in set $S$ is $S(S) \equiv n(\bmod d)$.
Let $T \subseteq S$, and $S(T) \equiv 0(\bmod d)$.
Then it can only be that $d=n, T=S$.
Therefore, when $d \geqslant n$, there do not exist four different sets, which does not meet the requirement.
When $d=n-1$, let
$a_{1} \equiv 0(\bmod d)$,
$a_{i} \equiv 1(\bmod d)(i=2,3, \cdots, n)$.
In this case, there only exist three different subsets $\left\{a_{1}\right\},\left\{a_{2}\right.$, $\left.a_{3}, \cdots, a_{n}\right\}$ and $S$ whose element sums are divisible by $d$, which does not meet the requirement.
Now consider the case when $d=n-2$.
First, prove the famous Erdös theorem as a lemma.

Lemma: Any $n$-element integer set has a non-empty subset whose element sum is divisible by $n$.
Proof: For the set $S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, consider $a_{1}, a_{1}+a_{2}, \cdots, a_{1}+a_{2}+\cdots+a_{n}$.
If one of these $n$ sums is divisible by $n$, then the lemma is proved. Otherwise, there exist $1 \leqslant i<j \leqslant n$, such that
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{i} \equiv a_{1}+a_{2}+\cdots+a_{j}(\bmod n) \\
\Rightarrow a_{i+1}+a_{i+2}+\cdots+a_{j} \equiv 0(\bmod n) .
\end{array}
$$

Thus, the set $T=\left\{a_{i+1}, a_{i+2}, \cdots, a_{j}\right\}$ meets the requirement.
The lemma is proved.
When $d=n-2$, by the lemma, there exists a non-empty subset $T_{1} \subseteq S \backslash\left\{a_{n-1}, a_{n}\right\}$ such that $d \mid S\left(T_{1}\right)$.

Take $a_{i} \in T_{1}$, by the lemma, there exists a non-empty subset $T_{2} \subseteq S \backslash\left\{a_{i}, a_{n}\right\}$ such that $d \mid S\left(T_{2}\right)$. Clearly, $T_{1} \neq T_{2}$.
(1) If $T_{1} \cap T_{2}=\varnothing$, let $T_{3}=T_{1} \cup T_{2}$.
Then $T_{3} \neq T_{1}, T_{2}$, and $d \mid S\left(T_{3}\right)$.
Take $a_{i} \in T_{1}, a_{j} \in T_{2}$, then $a_{i} \neq a_{j}$.
By the lemma, there exists a non-empty subset $T_{4} \subseteq S \backslash\left\{a_{i}, a_{j}\right\}$ such that $d \mid S\left(T_{4}\right)$.
Clearly, $T_{4} \neq T_{1}, T_{2}, T_{3}$.
Thus, $T_{1}, T_{2}, T_{3}, T_{4}$ are four different subsets that meet the requirement.
(2) If $T_{1} \cap T_{2} \neq \varnothing$, take $a_{r} \in T_{1} \cap T_{2}$.
By the lemma, there exists a non-empty subset $T_{3} \subseteq S \backslash\left\{a_{r}, a_{n}\right\}$ such that $d \mid S\left(T_{3}\right)$, and $T_{3} \neq T_{1}, T_{2}$.
Take $a_{s} \in T_{3}$, then $a_{s} \neq a_{r}$.
By the lemma, there exists a non-empty subset $T_{4} \subseteq S \backslash\left\{a_{r}, a_{s}\right\}$ such that $d \mid S\left(T_{4}\right)$.
Clearly, $T_{4} \neq T_{1}, T_{2}, T_{3}$.
Thus, $T_{1}, T_{2}, T_{3}, T_{4}$ are four different subsets that meet the requirement.
In summary, when $d=n-2$, the conclusion holds.
Therefore, $d_{\max }=n-2$.