Is it possible to stab (every rotation of) any four element subset of $\mathbb Z_n$ with less than $n/2$ elements? Say that $S\subset \mathbb Z_n$ is stabbed by $X\subset \mathbb Z_n$ if for every $t$ we have $(S+t)\cap X\ne \emptyset$.

Is there for every $|S|=4$ an $|X|<n/2$ that stabs it?

My motivation comes from here, where we would could improve the number of added edges.
In fact, instead of $\mathbb Z_n$ we could also use the dihedral group $D_n$ (where we pick some $t$ from $D_n$).
Update: This became a separate question.
I checked by hand, and it seems that there are no counterexamples for $n \le 8$, but I think that there should be for large enough $n$. (Update: After Seva's solution for $m\ge 5$, I'm not so confident about this anymore.)
More generally, instead of four, one can ask the following.

Is there an arbitrarily large $S\subset \mathbb Z_n$ that cannot be stabbed by any $|X|<n/2$?

 A: I think this is true for sets $S\subset\mathbb Z_n$ of size $m:=|S|\ge 5$. Here is a somewhat sketchy argument.
Let $\mu_m$ be the smallest value attained by the function $x+(1-x)^m$ on $[0,1]$. Numerical investigation shows that $\mu_m<1/2$, and this value is attained at a unique point $c_m\in(0,1/2)$. 
Choose $Y$ to be a random subset of $\mathbb Z_n$ of size $|Y|=c_m n$. For any fixed element $t\in\mathbb Z_n$, the probability that $Y$ is disjoint from $S+t$ is about $(1-c_m)^m$. Hence, the expected number of those $t$ with $(Y+t)\cap S=\varnothing$ is about $(1-c_m)^m n$. Choose some specific set $Y$ of size $|Y|=c_m n$ such that there are at most $(1-c_m)^m n$ those $t\in\mathbb Z_n$ with $(Y+t)\cap S=\varnothing$. For every such $t$, we can force $(Y+t)\cap S\ne \varnothing$ by expanding $Y$ by just one single element. As a result, we get a set $X$ of size $|X|\le c_mn+(1-c_m)^m n=\mu_m n<n/2$ such  that $(X+t)\cap S\ne \varnothing$ for any $t\in\mathbb Z_n$.
A: NEW VERSION: (What was I thinking?)
A greedy algorithm gives a stronger result.
THEOREM.  Consider any family $\mathcal F$ of $n$ 4-subsets of $\lbrace 1,\ldots,n\rbrace$. Then there is a set $X\subseteq \lbrace 1,\ldots,n\rbrace$ such that $|X|\lt\frac12 n$ and $X\cap S\ne\emptyset$ for all $S\in\mathcal F$.
PROOF.
Let's choose $X$ one element at a time.  If we have hit $h$ of the sets after $|X|=k$ elements, by averaging there is some $x\in \mathbb{Z}-X$ that hits at least 
$$\left\lceil \frac{4(n-h)}{n-k} \right\rceil$$
additional sets. Put $x$ into $X$ and continue until all sets have been hit.
The progress is always bounded like this (ignoring rounding to integer):  For $\frac{1}{13}n$ steps, 4 subsets are hit at each step. Then for $\frac{6}{65}n$ steps, 3 subsets are hit at each step. Then for $\frac{54}{455}n$ steps, 2 subsets are hit at each step. Then the remaining $\frac{81}{455}n$ subsets are hit individually.
This gives $\frac{212}{455}n$ steps altogether. Properly rounding the phase lengths to integer can't increase the value by more than 3, so it suffices to test small $n$ to achieve $\frac12 n$.
In fact it seems that $\frac{212}{455}n$ is an upper bound for all $n$.  It is achieved when $n$ is a multiple of 455.
For $4\le n\le 10^7$, the solution to the recurrence is either $\bigl\lfloor\frac{212}{455}n\bigr\rfloor-1$ or $\bigl\lfloor\frac{212}{455}n\bigr\rfloor$, with a non-obvious pattern.
A: I think this is true for any group $G$, and if $|G|$ is odd then $|S| \geq 3$ suffices. I will use multiplicative notation instead of additive notation since I do not assume that $G$ is commutative.
Theorem. If $G$ is a finite group and $S \subseteq G$ with $|S| \geq 4$ then there is a set $X \subseteq G$ such that $|X|<|G|/2$ and $gS \cap X \neq \varnothing$ for every $g \in G$. If $|G|$ is odd then $|S| \geq 3$ suffices.
Translating $S$ if necessary, we may assume that $1 \in S$. Let $H$ be the subgroup generated by $S$. We will first find a stabbing set for $S$ in $H$: a set $X \subseteq H$ with $|X| < |H|/2$ such that $hS \cap X \neq \varnothing$ for every $h \in H$. Then $g_1X \cup \cdots \cup g_kX$ will be as required where $g_1,\ldots,g_k$ are representatives for the left cosets of $H$ in $G$. (If $g \in G$ then $g = g_ih$ for some $i \in \{1,\ldots,k\}$ and some $h \in H$. If $x \in hS \cap X$ then $g_ix \in g_ihS \cap g_iX = gS \cap g_iX$.)
Let $S_0 = S-\{1\}$ and define the directed graph $D$ with vertex set $H$ and directed edges $(hs,h)$ for $s \in S_0$. This is the Cayley digraph for the generating set $S_0^{-1}$. Note that this is a strongly connected directed graph. In fact, $D$ is always strongly $\lceil |S|/2 \rceil$-connected by a result of Hamidoune:
Hamidoune, Yahya Ould, Quelques problèmes de connexité dans les graphes orientés, J. Comb. Theory, Ser. B 30, 1-10 (1981). ZBL0475.05039.
In the case of interest where $|S|\geq4$, this only ensures that $D$ is strongly $2$-connected. However, it turns out that $D$ is $3$-connected except in one scenario, which can be dealt with separately.
Let's first deal with the case where $D$ is strongly $3$-connected.
Lemma. There is a directed spanning tree $T$ for $D$ where in the root $1$ is connected to two leaves of $T$.
Proof: Pick distinct $s,t \in S_0$. Remove $s^{-1},t^{-1}$ from $D$ and form a directed spanning tree $T_0$ rooted at $1$ for the rest. (You can find one of these greedily since $D-\{s^{-1},t^{-1}\}$ is strongly connected.) Then add the edges $(1,s^{-1})$ and $(1,t^{-1})$ to $T_0$ to obtain the desired tree $T$. $\square$
Paint elements of $H$ red and blue starting from the leaves of $T$ as follows. Once all the immediate successors of $h$ in $T$ have been painted red or blue, paint $h$ blue if at least one successor of $h$ has been painted red, otherwise paint $h$ red. So the leaves of $T$ are all red, their immediate predecessors are all blue, then the nodes whose immediate successors are either leaves or immediate predecessors of leaves are determined, and so on all the way down to the root $1$ of $T$.
Let $X$ be the set of blue nodes and let $Y$ be the set of red nodes. Note that $1 \in X$.
Claim 1. $X$ is stabbing for $S$.
Proof: Given $h \in H$. If $h \in X$ then $h \in hS \cap X$. If $h \in Y$ then $h$ has an immediate predecessor in $T$, which is of the form $hs$ for some $s \in S_0$ and must be painted blue. Therefore $hs \in hS \cap X$.
Claim 2. $|X| < |H|/2$.
Proof: The map which sends every red element to its immediate predecessor is a surjection from $Y$ onto $X$. This is not a bijection because $1$ is a blue node attached to two leaves, which are both red. Therefore $|X|<|Y|$ and since $H = X  \cup Y$, $X \cap Y = \varnothing$ it follows that $|X| < |H|/2$.
If $D$ is merely known to be strongly $2$-connected, which follows from $|S| \geq 3$, then we can find a spanning tree $T$ for $D$ where the root $1$ is connected to at least one leaf. This ensures that $1$ is colored blue to make Claim 1 work. The surjection from Claim 2 could be a bijection but we still have that $|X|\leq|Y|$. If $G$ and therefore $H$ have odd order, we can't have $|X|=|Y|$ and Claim 2 still holds. So $|S|\geq3$ suffices assuming that $|G|$ is odd.
Assuming $|S|=4$, the one scenario where $D$ is not strongly $3$-connected is when $S = \{1,a,b,ba\}$ and $a^2 = 1$ (up to translation). This follows from Hamidoune's atom theory, which says that the connectivity of the Cayley graph $D$ of $S_0^{-1}$ in $H$ can be measured by an atom $A$ which is a subgroup of $H$ generated by $A \cap S_0^{-1}$ such that $AS_0^{-1} \setminus A$ is a minimal cut set of $D$. 
Hamidoune, Yahya Ould, On the connectivity of Cayley digraphs, Eur. J. Comb. 5, 309-312 (1984). ZBL0561.05028.
Specializing to the case at hand where $|S|=4$ and $D$ is strongly $2$-connected but not strongly $3$-connected, we see that the only possibilitiy is that $S_0 = \{a,b,ab\}$ where $a^2 = 1$ and $A = \{1,a\}$.
Let $X \subseteq H$ select one element from each left coset of the subgroup $\{1,a\}$ in $H$, except for the coset $\{b^{-1},b^{-1}a\}$. Without loss of generality $1 \in X$. Thus $|X| = |H|/2 - 1$. To see that this works, given $h$ in $H$ there are two cases:


*

*If $h \in \{b^{-1},b^{-1}a\}$ then $1 = b^{-1}b = b^{-1}aab \in hS \cap X$.

*Otherwise, $X$ contains one of $hA = \{h,ha\}$. Since $hA \subseteq hS$ we see that $hS \cap X \neq \varnothing$.
