TITLE: Is there a nice way to characterize this subset of $Z_n$?
QUESTION [12 upvotes]: This is a problem I played around with several years ago.  I proved a few things, but the proofs were long and messy and not worth reproducing here.  
Given a positive integer $n$ and a set $S$ such that $S \subset Z_n$, define a focal point of $n$ and $S$ to be any $a \in Z_n$ such that for each $x \in S$, there exists $y \in S$ such that $x+y \equiv a \mod n$.  It is possible to have $x = y$.  
Example 1. $n=15$, $S_1=\{1,2,6,7,10,11,12,13\}$.  Then $n$ and $S_1$ have the focal point $8$ because $1+7 = 2+6 = 10+13 = 11+12 = 8$.  By exhaustion one can show that there are no others.  
Example 2: $n=15$, $S_2=\{1,2,6,7,11,12,13\}$.  These have no focal points, even though $S_2$ is simply $S_1$ with $10$ removed.  In fact, removing any element from $S_1$ results in a set that has no focal points with $n$.  
Example 3: $n = 15$, $S_3=\{1,2,6,7,11,12\}$.  $S_3$ is obtained by removing both $10$ and $13$ from $S_1$.  $S_3$ and $n$ have 3 focal points: $3, 8$, and $13$.
Let $F(n,S)$ be the set of focal points of $n$ and $S$.  My questions are these:


What can we say about the size of $F(n,S)$?  I was able to prove that it must be $0$ or a divisor of $n$, but surely one can say more.
Even better, is there a nice way to characterize $F(n,S)$?


My hunch is that the right algebraic structure will make what's going on here crystal clear.  Unfortunately, my algebra is quite rusty.
Added: One thing to observe is that in the case of Example 3, $S_3$ splits nicely into residue classes $\bmod 5$: $\{1, 6, 11\}$ and $\{2, 7, 12\}$.  If $n_4 = 5$ and $S_4 = \{1,2\}$, then clearly $n_4$ and $S_4$ have one focal point: $3$.  Surely it's not a coincidence, then, that $n_3 = 15$ and $S_3=\{1,2,6,7,11,12\}$ have 3 focal points.  So perhaps there's a characterization that has to do with when $S$ can be divided into residue classes modulo a divisor of $n$?

REPLY [2 votes]: Two simple remarks.
First, $|F(n,S)| \leq |S|$ for obvious reasons.
Second, suppose $a,b \in F(n,S)$. Thus $S$ is closed under $x \mapsto a-x$ and $x \mapsto b-x$, so it is closed under their composition, which is the shift $x \mapsto a-b+x$. Similarly, if $a,b,c \in F(n,S)$ then $a-b+c \in F(n,S)$. So $F(n,S)$ is also closed under shift by $a-b$.
Edit: Here's a nicer way to define $F(n,S)$. Let $T = \{ x : -x \in S \}$. Then $a \in F(n,S)$ iff $S$ is equal to the $a$th cyclic shift of $T$.
Another remark: If $a \in F(n,S)$ is even then $S$ is symmetric with respect to $a/2$. I guess this is also true for odd $a$. In order to see this, it is perhaps better to replace $S$ with $n\mathbb{N} + S$, and then $F(n,S)$ consists of all points of symmetry (doubled).