Separating $p-1$ non-multiples of odd prime $p$ into two groups of the same sum in terms of $\pmod p$ 
Show that it is possible to separate $p-1$ non-multiples of odd prime $p$ into two groups of the same sum in terms of $\pmod p$.

Something similar has been posted about a year ago but got no answers. I suppose it is here. Now my approach is induction, not on $p$ of course, but on how many numbers given in the beginning.
Call the given numbers $a_1,a_2,\ldots$.First change the problem to the $+,-$ language like in the link. That is $$\exists\epsilon_i\in\{-1,1\},i\in[1,p-1]\mbox{ so that }p\mid\sum_{k=1}^{p-1}\epsilon_ka_k.$$I want to show:$$\forall s\in [1,p-1],\mbox{there’re at least }s+1\mbox{ kinds of remainders}\sum_{k=1}^s\epsilon_ka_k\pmod p.$$ If so, let $s=p-1$ and the problem gets solved.
 A: I've been chipping away at this slowly for a couple of days, and wanted to post a partial result. I believe the question of "Is this true only of primes?" has already been answered, but here's an elementary proof not using the Cauchy-Davenport Theorem, as well as some introductory stuff and further thoughts. I'm still chipping away at the central question, though.

I'll go ahead and assume that the residues need not be unique, and that the empty set is allowed. For now, we will consider the possibility that these conditions can be satisfied for composite numbers. Also, read "set" as almost always being equivalent to "multiset."
Let $A$ be a multiset $\{a_1, a_2, \cdots a_{n-1}\}$ where $a_i$ are nonzero residues modulo $n>2$. Let $S_A$ be the sum modulo $n$ of the residues in the multiset (and use the same notation for other sets or multisets). We desire a partition of $A$ into disjoint submultisets $B$ and $C$ such that $S_B = S_C$. If $C=\emptyset$, then $S_C=0$.
To have $S_B = S_C$, both sums must be equal to half the original sum: $S_B = S_C = \frac12 S_A \pmod n = T$. ($T$ is the "target sum" for $A$.) Equivalently, we can write $S_B-S_C=0$, as you pointed out above.
Let $\mathbf{H(n)}$ be the following hypothesis: In modulo $n$, for all multisets $A$ that can be constructed as above:
$$(\exists B \exists C)(A = B \cup C, B \cap C = \emptyset, S_B \equiv S_C \equiv \textstyle{\frac12} S_A \pmod n)$$

Can $H(n)$ be true if $n$ is not prime? It's trivial to show $H(n)$ cannot be true if $n$ is even: if $S_A$ is odd, $T = \frac12 S_A$ is indeterminate, as $2$ is not invertible in even moduli. What about odd semiprimes or odd prime powers? We can prove by counterexample that $H(n)$ is not true for, in fact, all odd composite numbers:
Create $A$ containing $n-2$ copies of a residue $m$ that is a factor of $n$, and a single $1$. (For example, if $n=9$, we could use $A = \{3,3,3,3,3,3,3,1\}$.) For this multiset:
$$S_A \equiv m(n-2) + 1 \equiv mn-2m+1 \equiv 1-2m \pmod n$$
Hence $T \equiv \frac12 S_A \equiv (n+1)/2 -m \pmod n$.  Meanwhile, the partial sums $S_B$ or $S_C$ that can be created from $A$ are all congruent to $km$ or $km+1$ (where $k \le n-2)$. If $H(n)$ were true, one of these would be true:
$$
\begin{gather}
km \equiv (n+1-2m)/2 \implies 2km \equiv n+1-2m \implies (2k+2)m \equiv n+1 \\
km+1 \equiv (n+1-2m)/2 \implies 2km+2 \equiv n+1-2m \implies (2k+4)m \equiv n+1
\end{gather}$$
Now recall that $m$ is a factor of $n$, meaning $(m,n) = m$ and so $(qm, n) = m$. But that means that $qm \ne n+1$, which implies that neither of those equivalences is true.
This construction works for all factors of all odd composite numbers; therefore, $n$ must be prime after all. We shall use $H(p)$ henceforth.

There are some trivial cases for some $A \pmod p$:

*

*If $S_A \equiv 0 \pmod p$, then $B = A$ and $C = \emptyset$. (This is the case if all elements are unique modulo $p$.)

*If some element $a_i \equiv T \pmod p$, then $B = A \setminus \{a_i\}$ and $C = \{a_i\}$

*If all elements of $A$ are congruent modulo $p$, then $B=C$, with each containing half the elements of $A$.

Interestingly, $H(3)$ is shown true with only these trivial cases. There are only three $A$, as follows. $A=\{1,2\}$ has the two unique residues modulo $3$, which falls under Case 1. $A=\{1,1\}$ and $A=\{2,2\}$ have both residues congruent and fall under Case 3.

What do we know about sets that aren't trivial? We know there are no elements $a_i \equiv 0 \pmod p$, and no $a_i \equiv T \pmod p$. That implies that at least two elements must be congruent modulo $p$. So we can safely claim that for all $A$, $a_{p-1} \equiv a_{p-2} \pmod p$.

(More to come, I promise.)
