In a now-deleted question (previously here), user Lục Trường Phát posted what I thought was an interesting problem from a math competition at their school. To be clear, I don't know anything about this context, only that I found the question interesting. Let me rephrase it slightly:
Over all packings of $n$ cliques into the complete graph $K_n$ (that is, using each edge in at most one clique), if the $k^\text{th}$ clique has $a_k$ vertices, what is the maximum value of $a_1+a_2+\cdots + a_n$?
After playing with this for a while and in particular finding some optimal packings for $n\leq 8$, I've conjectured that among the optimal packings is one such that all of the cliques have either $c$ or $c+1$ vertices, for $c=\left\lfloor\frac{1+\sqrt{4n-3}}{2}\right\rfloor$. This choice of $c$ is certainly the largest possible, by the natural edge-counting argument.
This would give fairly tight bounds on the maximum sum, including an exact answer when $\sqrt{4n-3}$ is an (odd) integer, but so far my attempts to prove it haven't succeeded.