Take an abelian group $(\mathbb{Z}_n,+)$ and enumerate all partitions of two elements (i.e. $x=x_1+x_2$) of each element $\{0,1,...,n-1\}=\mathbb{Z}_n$. Take, for example, abelian groups $\mathbb{Z}_9$ and $\mathbb{Z}_{10}$. Now, the enumerations of the partitions of these groups would be the following (note that $x=x_1+x_2$ where $x_1=x_2$ is not allowed, i.e. $x_1$ and $x_2$ may not be equal!):
$\mathbb{Z}_9:$
$0=8+1,7+2,6+3,5+4$
$1=8+2,7+3,6+4,0+1$
$2=8+3,7+4,6+5,0+2$
$3=8+4,7+5,0+3,1+2$
$4=8+5,7+6,0+4,1+3$
$5=8+6,0+5,1+4,2+3$
$6=8+7,0+6,1+5,2+4$
$7=0+7,1+6,2+5,3+4$
$8=0+8,1+7,2+6,3+5$
$\mathbb{Z}_{10}:$
$0=9+1,8+2,7+3,6+4$
$1=9+2,8+3,7+4,6+5,0+1$
$2=9+3,8+4,7+5,0+2$
$3=9+4,8+5,7+6,0+3,1+2$
$4=9+5,8+6,0+4,1+3$
$5=9+6,8+7,0+5,1+4,2+3$
$6=9+7,0+6,1+5,2+4$
$7=9+8,0+7,1+6,2+5,3+4$
$8=0+8,1+7,2+6,3+5$
$9=0+9,1+8,2+7,3+6,4+5$
Now, my question is the following: how many ways can we choose the total of $\lfloor \frac{n-1}{2}\rfloor$ distinct elements from $\mathbb{Z}_n$ such that when these elements are deleted, there exists at least one remaining element in $\mathbb{Z}_n$ that may not be expressed as a partition $x=x_1+x_2$ anymore. For instance, if we look at $\mathbb{Z}_9$ then $\lfloor \frac{9-1}{2} \rfloor=4$ and if we delete $\{8,2,3,5\}$ from $\mathbb{Z}_9$ then we cannot express $0$ as a partition $0=x_1+x_2$ where $x_1,x_2\in \mathbb{Z}_9 \setminus \{8,2,3,5\}$.
I noticed that no $\lfloor \frac{n-1}{2} \rfloor$ consecutive elements may be deleted from $\mathbb{Z}_n$ so there must be at least $n$ different ways to do this. However, there have to be more than $n$ due to, for instance, my previous example of $\mathbb{Z}_9\setminus \{8,2,3,5\}$.