Nice double-counting problem! The answer is 1012, for $2n$ people the answer is $n$.

Here is a really simple approach. The problem is equivalent to the following problem in Graph Theory. Problem: Let $G$ be a simple graph with 2024 vertexes. $d_1\le d_2\le .....\le d_{2024}$ are the degrees of the $2024$ vertexes. Then we have: $\displaystyle \sum_{i=1}^{2024} |d_i-(i-1)|\ge 1012$ Solution: (Solved with Chunji Wang (2023 IMO CHN1)) In fact, $\displaystyle \sum_{i=1}^{2024} |d_i-(i-1)| \ge (1012+1013+......+2024)-(0+1+......+1011) +\displaystyle \sum_{i=1}^{1012}d_i-\displaystyle \sum_{j=1013}^{2024}d_j \ge 1012^2-2*E(K_{1012})=1012$ Construction: In graph $G$, there exists an edge between $v_i$ and$v_j$ IFF. $i+j\ge2025$