The wording is a bit iffy for those who know graph theory. Here, by "path" they mean the graph-theoretic edge - not path. Let's say we have a point $ Q_{n+1}$ in a set $ S_n$ of points connected by an edge to points $ Q_1,Q_2,...,Q_n$ and we want to determine if $ Q_{n+1}$ is connected to another vertex in $ S_n$. Then we simply have to check whether $ 1991-|S_n|-n< 1593$. Then we are assured that there exists a set $ S_{n+1}\in S_n$ with $ |S_{n+1}|\ge|S_n|+n+1593-1991=|S_n|+n-398$ and for all $ P\in S_{n+1}$, $ P$ is connected by an edge to points $ Q_1,Q_2,...,Q_{n+1}$. Since the sum of the degrees of a graph must be even, we must have one vertex with degree at least $ 1594$. It follows that $ |S_1|\ge 1594$. Then we have $ |S_2|\ge1197,|S_3|\ge801,|S_4|\ge406,|S_5|\ge12$. Since $ 12>5$, there exists $ Q_6\in S_5$ so that $ Q_1,Q_2,Q_3,Q_4,Q_5,Q_6$ are our desired points.

We use Zarankiewicz Lemma. Assume the graph is $6$-free. Then by Zarankiewicz lemma, there exists a vertex with at most degree $[\frac{4}{5}*1991] = 1592$. However, this is a contradiction as all vertices have degree strictly greater than $1592$.

va2010 wrote: We use Zarankiewicz Lemma. Assume the graph is $6$-free. Then by Zarankiewicz lemma, there exists a vertex with at most degree $[\frac{4}{5}*1991] = 1592$. However, this is a contradiction as all vertices have degree strictly greater than $1592$. In the actual contest though, (Indian National MO (INMO) in my case), would we need to prove the Zarankiewicz Lemma, or would we get full marks for just this much?