Is there a nice way to characterize this subset of $Z_n$? 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:

  
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*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 $n$? 
 A: Not an answer, but an attempt to supply an algebraic structure. A regular $n$-sided polygon has $2n$ symmetries - $n$ rotations (counting the "do-nothing" rotation), and $n$ flips. Now take your $n$-sided polygon and color in the vertices corresponding to your set, $S$. I think you'll find the number of focal points is the number of flips that are symmetries of this colored polygon, or, to put it another way, the number of axes of symmetry. 
A: 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).
