Let's bump this. Here some thoughts on this nice graph problem: Take the maximal matching $M$ with $2n$ vertices and $n$ edges. Suppose FTSoC $n<=150$, so the remaining vertices (say they form a set of vertices $N$) form an anticlique (due to the maximality) with size at least $150$. But let's note that $N$ has even size, and due to the condition that $|N|<=199$, we have $150<=|N|<=198$. Also, note that for each edge $v_iw_i$ in $M$, at most one of the endpoint has neighbors in $N$, otherwise we get contradiction with the maximality, except for the case when each of them has exactly one neighbor in $N$, and this neighbor is common (so we have a triangle). In addition, I was also thinking of the following: if we take $50$ (or eventually less) vertices of the vertices of $M$, and add them to the anticlique $N$, we shall have at least one edge in this set, and in addition, if we take this $50$ vertices so that they don't have neighbors in $N$, then we have that there will be at least one edge among them. Lastly, I noted that we let WLOG $v_i$ be the vertices of $M$ which have at most one neighbor in $N$ (in particular, we could take some of those in the above step), then each such $v_i$ has at least $99$ neighbors in $M$, including its neighbor $w_i$.

A very nice problem, especially it is only using some basic ideas and techniques... First consider the obvious graph theory interpretation and consider graph $\mathcal{G}$ , and assume for the sake of contradiction that in chosen $302$ vertices we can't find a matching of size $151$ in that $302$ vertices. Call the isolated part of that $302$ vertices be the set of vertices in that $302$ vertices which can't managed to be pair up with others. Now take a set of $302$ vertices $\mathcal{M}$ ,such that it's isolated part $\mathcal{I}$ is of minimum size. Note that none of vertices in $\mathcal{I}$ are connected to itself or to $\mathcal{G} - \mathcal{M}$ , otherwise we can get another set of $302$ vertices with smaller size of its Isolated part, which is a contradiction to the minimality of $\mathcal{I}$ , also note that |$\mathcal{I}|$ must be even (say = $2k$), so now as $\mathcal{G} - \mathcal{M}$ has $148$ vertices, so $148 +2k \le 199$ , giving $k \le 25$ , but note that in $\mathcal{G} - \mathcal{M}$ there are $25$ pairs which can be sent into kayaks, and $\mathcal{M} - \mathcal{I}$ , contains $151-k$ pairs, so $\mathcal{G} - \mathcal{I}$ contains $= 151-k +25 \ge 151$ , so from $\mathcal{G} - \mathcal{I}$ , choose any $151$ pairs, which is required contradiction. Hence we are done $\blacksquare$

could anyone plz check my above solution..

It's probably correct, but why there are at least $25$ edges in $G-M$ (I'm being dumb now probably, sorry if this is the case)? Except that, everything else seems ok. Edit: Wait, actually $G-M$ should be anticlique (if it is not, you take and edge from it and replace a pair of non-adjacent vertices in $I$ with it, and thus $I$ becomes smaller) , you can't have edges there I think.

I will prove that the bound can be improved to $176$ and that it cannot be improved anymore. Let $k\leq175$ pairings be the maximum we can have and observe a graph-theoretic configuration (possibly one of more of them) where that maximum is achieved. Call a set of vertices in pairs $M$ and the set of remaining vertices $N$.

. For pairs in group $c$ note that the vertices which are not connected to any vertex in $N$ cannot be connected with each other, otherwise we could dismantle 2 pairs and obtain 3 new pairs, contradicting maximality. (1) Obviously, no two vertices in $N$ are connected and $|N| \geq 100$ (cardinalities of groups of pairs refer to number of pairs in those groups and cardinality of $N$ refers to number of vertices in $N$). (2) Claim. $|c| \geq 100$

However, if we observe $200$ vertices ($100$ from the group $c$ which are not connected to any vertex in $N$ and $100$ from $N$) , the condition that for any $200$ vertices there is a connection among them is not satisfied (from (1) and (2)), yielding the desired contradiction.