, 2015

$d \ge n$ is not possible. To see this, take a set $S$ of $n$ integers so that each element of $S$ is equal to $1 \pmod{n}$. The sum of any nonempty subset $T$ of $S$ is equal to $\#T \pmod{d}$. Since $d \ge n$, the only possibility for this to hold is if $d=n$ and $\#T=n$, i.e., $T=S$. This proves that $d \ge n$ fails the given condition.

$d=n-1$ is not possible. In this case, take a set $S = \{a_1, \dots, a_n\}$ of $n$ integers so that $a_1 = 0$ and $a_i = 1 \pmod{d}$ if $1 < i \le n$. The only subsets of $S$ whose sums are divisible by $d$ are:
$$\{a_1\}, \{a_2, \dots, a_n\} \quad \text{and} \quad S.$$
We will show that $d=n-2$ satisfies the given condition. So the largest $d$ is $n-2$.

Lemma. (Erdös) Any set of $n$ integers has nonempty subset whose sum is divisible by $n$.

Consider a set of $n$ integers $\{a_1, \dots, a_n\}$. Consider the set of $n$ numbers $\{a_1, a_1 + a_2, \dots, a_1 + \dots + a_n\}$. If one of these numbers is equal to $0 \pmod{n}$, then we are done. Otherwise, there are two of them that are equal (mod $n$). Say
$$a_1 + \dots + a_i = a_1 + \dots + a_j \pmod{n}, i < j.$$
Then $a_{i+1} + \dots + a_j = 0 \pmod{n}$.

We return to the proof that $n-2$ satisfies the given condition. Let $S = \{a_1, \dots, a_n\}$ be a set of $n$ integers. By the lemma, there is a subset $T_1$ of $\{a_1, \dots, a_{n-2}\}$ whose sum is divisible by $n-2$. We may also assume that $T_1$ is minimal in the sense that removal of any element from $T_1$ would either render it empty, or make the sum of its elements indivisible by $n-2$. Take $a_i \in T_1$. Choose nonempty subset $T_2$ of $S \setminus \{a_i, a_{n-1}\}$ whose sum is divisible by $n-2$. Clearly, $T_1 \neq T_2$.

If $T_1 \cap T_2 = \emptyset$, pick $a_i \in T_1, a_j \in T_2$ and let $I = S \setminus \{a_i, a_j\}$. By the lemma, there exists a subset $T_3$ of $I$ whose sum is divisible by $n-2$. Clearly, $T_1, T_2, T_1 \cup T_2$ and $T_3$ are four distinct sets, the sum of the elements of each of which is divisible by $n-2$.

The previous paragraph shows that if there are two disjoint nonempty subsets of $S$ whose sums are both divisible by $n-2$, then we are done. Assume that this does not occur. Let $T_1$ and $T_2$ be two distinct subsets of $S$ whose sums are divisible by $n-2$. Choose $a_i \neq a_j$ so that $a_i \in T_1$ and $a_j \in T_2$. By the Lemma, there exists a subset $T_3$ of $S \setminus \{a_i, a_j\}$ whose sum is divisible by $n-2$. In particular, $T_3$ is different from both $T_1$ and $T_2$. Moreover, by the previous paragraph, we may assume that $T_i \cap T_j \neq \emptyset$ if $1 \le i < j \le 3$. Then we can choose two distinct points $a_r$ and $a_s$ so that $\{a_r, a_s\} \cap T_i \neq \emptyset$ for $i=1, 2, 3$. Use the lemma again to find $T_4 \subseteq S \setminus \{a_r, a_s\}$ so that the sum of the elements in $T_4$ is divisible by $n-2$. By construction $T_4$ is different from $T_1, T_2$ and $T_3$.