Let $n$ an positive integer, $S\subset\mathbb{Z}$ a set of size $n-1$, satisfying that for any non-empty set $T$ which is subset of $S$, sum of elements of $T$ is not divisible by $n$. Can we prove that every element of $S$ has the same value modulo $n$?
In maths symbol, $|S|=n-1, \forall\emptyset\ne T\subseteq S, n\nmid\sum_{x\in T}x\Rightarrow\forall x,y\in S, x\equiv y\pmod n$
It can be easily shown by pigeonhole principle that every integer set of size $n$ has a non-empty subset whose sum of elements is divisible by $n$, discussed here.
My work so far:
I wrote a program and found that for $n\le10$, my guess is right.
We can discuss the question in modulo $n$ arithmetic. Let $x$ be the count of $1$ in $S$. Then there is no elements $n-i(1\le i\le x)$, otherwise adding $1$ will get a sum of $n$. And sum of elements ranging in $[2,x+1]$ is less than $n-x$, otherwise we can begin with all elements in $[2,x+1]$, remove elements one by one and adjust the sum of elements into $[n-x,n]$, which constructs a subset of sum $n$ with the help of $1$. However, I have no idea about numbers in $[x+2,n-x-1]$.
