If I have a set, I know it has a subset whose sum is divisible by the original set's cardinal. (See here)
If I have a set where none of its proper subsets' sums is divisible by the original set's cardinal, does all of the original set's elements have the same reminder modulo the original set's cardinal?
It is true that all numbers in the set are congruent modulo the cardinality of the set.
Denote the set by $S$ and suppose $|S| = n \geqslant 2$. If $n = 2$, suppose $S = \{a, b\}$, then $2 \not\mid a$, $2 \not\mid b$ implies $a \equiv b \equiv 1 \pmod{2}$.
For $n \geqslant 3$, if $a, b \in S$ satisfy $a \not\equiv b \pmod{n}$, suppose $S = \{a, b, c_1, \cdots, c_{n - 2}\}$. Now consider the following $n$ proper subsets:$$ \{a\}, \{b\}, \{a, b\}, \{a, b, c_1\}, \cdots, \{a, b, c_1, \cdots, c_{n - 3}\}. $$ Because for any of these sets, $n$ does not divide the sum of its elements, by the pigeonhole principle, there are two sets $A$ and $B$ such that the sum of $A$'s elements is congruent to that of $B$'s. Note that if $A = \{a\}$, then $B \neq \{b\}$, and if $A = \{b\}$, then $B \neq \{a\}$, therefore either $A$ and $B$ are in this chain:$$ \{a\} \subset \{a, b\} \subset \{a, b, c_1\} \subset \cdots \subset \{a, b, c_1, \cdots, c_{n - 3}\}, $$ or in this chain:$$ \{b\} \subset \{a, b\} \subset \{a, b, c_1\} \subset \cdots \subset \{a, b, c_1, \cdots, c_{n - 3}\}. $$ Without loss of generality, suppose $A \subset B$, then $B \setminus A$ is a non-empty proper set of $S$ but $n$ divides the sum of the elements of $B \setminus A$, a contradiction.
Therefore, all elements in $S$ are congruent modulo $|S|$.
