, 2020

Solution:
Assume for the sake of contradiction that there exist no such $A, B$. Pair up each subset with its complement, like so:
$$\begin{aligned} \{1,2,3, \ldots, n\} & \leftrightarrow \{n+1, n+2, \ldots, 2n\} \\ \{1,2,3, \ldots, n-1, n+1\} & \leftrightarrow \{n, n+2, \ldots, 2n\} \\ \{1,2,3, \ldots, n-1, n+2\} & \leftrightarrow \{n, n+1, n+3, \ldots, 2n\} \\ & \vdots \end{aligned}$$
Note that for each pair, we can have at most one of the two in $S$. Since $S$ has $\frac{1}{2}$ of the total number of subsets with size $n$, it must be that we have exactly one element from each pair in $S$. For any $s_{0} \in S$, none of the subsets that share exactly one element with $s_{0}$ can be in $S$, so their complements must be in $S$. This means that every subset with $n-1$ shared elements with $s_{0}$ must be in $S$. Without loss of generality, assume $\{1,2,3, \ldots, n\} \in S$. Then, $\{2,3,4, \ldots, n+1\} \in S$, so $\{3,4,5, \ldots, n+2\} \in S$. Continuing on in this manner, we eventually reach $\{n+1, n+2, \ldots, 2n\} \in S$, contradiction.

Solution:
Let $[2n] = \{1,2, \ldots, 2n\}$. Consider the following cycle of $2n-1$ sets such that any two adjacent sets have an intersection of size $1$:
$$\begin{aligned} & \{1,2,3, \ldots, n\} \\ & \{1, n+1, n+2, \ldots, 2n-1\} \\ & \{1, 2n, 2,3, \ldots, n-1\} \\ & \vdots \\ & \{1, n+2, n+3, \ldots, 2n\} \end{aligned}$$
If $S$ contains two adjacent elements of the cycle then we are done. For each permutation $\sigma$ of $[2n]$, we can consider the cycle $C_{\sigma}$ of sets obtained after applying $\sigma$, i.e. $\{\sigma(1), \sigma(2), \sigma(3), \ldots, \sigma(n)\}$ and so on. In total, each subset of $[2n]$ with size $n$ appears $(2n-1)(n!)^{2}$ times across all the cycles $C_{\sigma}$, so
$$\sum_{\sigma} |C_{\sigma} \cap S| = |S| (2n-1)(n!)^{2} = \frac{2n-1}{2} (2n)!$$
where the sum is over all $(2n)!$ permutations of $[2n]$. This means that on average across possible cycles, $\frac{2n-1}{2}$ of its elements are in $S$. Thus, if we select a cycle $C_{\sigma}$ uniformly at random, with positive probability we will have $|C_{\sigma} \cap S| \geq n$, so two adjacent elements in this cycle will be in $S$. Therefore, there must exist some two subsets in $S$ that share at most one element.

This proof will work under the weaker condition $|S| > \frac{n-1}{2n-1} \binom{2n}{n}$.

Remark. A family of sets such that $|A \cap B| \geq t$ for every pair of distinct sets $A, B$ is called $t$-intersecting. Ahlswede and Khachatrian solved the problem of determining the largest $k$-uniform $t$-intersecting family. See "Katona's Intersection Theorem: Four Proofs" or "The Complete Intersection Theorem for Systems of Finite Sets" for exact results.