Is Finding A Hitting Set of Size n/2 NP-Hard?
I think the easiest way is to reduce 


  $ VC_{n/2}=\{G∣G=(V,E) $ has a vertex cover of size $|V(G)|/2\}.$ ( https://cs.stackexchange.com/a/90487/72705) 


to

$ HS_{n/2}=\{(S,F)∣\exists H\subseteq S: $$H$ is a HS of $F$ of size at most $|S|/2 $$\}$.

In  https://cs.stackexchange.com/a/90487/72705 , they show that $VC_{n/2}$ is $NP$-hard.

Now, we reduce $VC_{n/2}$ to $HS_{n/2}$, which is the standard reduction from normal $VC$ to $HS$:


$S = V$
$F = E$, where $E:= \{e_1,e_2,\ldots,e_m\}$ with $e_i = \{v_j, v_k\}$ and $v_j, v_k \in V$


Clearly, this reduction can be done in poly. time and is correct, as one element in $F$ corresponds to one edge in $E$, so the HS can be mapped one to one to the vertex cover set and vice versa.  

As you can see, the vertex cover set has the same number of elements as the corresponding hitting set (if there exists one). Also, $|S| = |V|$. Therefore, the Vertex Cover Set of size $|V|/2$ gives a HS of size $|S|/2$. To get a HS of size $(|S|/2) -1$, one could add two completely new elements $\alpha, \beta$ to $S$ (let this be $S'$) without changing $F$. Therefore, $\alpha$ and $\beta$ will never be in the HS of $F$, but the Hitting Set now has a size of $|S|/2 = (|S'| -2) / 2 = (|S'|/2) -1$.