Question: Let $n$ be some odd, positive integer and $S = \{1,2,\dots, n\}$. Show that if $\frac {n+3}2 $ numbers are selected from $S$, then there will be distinct numbers $x,y$ in that selection such that $x+y = n+1$.
NOTE: Not about probability, simply want to guarantee. Furthermore, this is a pigeonhole problem (I suspect) so I would like to solve it in that manner.
Let $H_j = \{n - j + 1, j\}$ for $1\leq j \leq \frac{n-1}{2}$ be our pigeonholes. But note that this does not form a partition of $S$ yet. Let $k = 1$, be the missing middle element in the set due to odd cardinality. Thus, we need only pick $\frac{n-1}2 + k + 1$ pigeons. Subbing $k$ value, we have $\frac{n-1}2 + 1 + 1 = \frac{n + 3}{2}$.
Note, I conveniently picked $k = 1$ to represent middle element even though it might not even get picked in the selection. But you get the point, right? In addition, I noticed the equation $x+y = n+1$ was drawing inspiration from Gauss's famous formula for summing the set $\{1, \dots, n\}$ where $n=100$ (or something of that nature).