(I wrote it by solving it in my head, probably the wrong solution, I will try again with pen and paper.) Claim:For $2n$ veritces at least $2n-1$ days are required From our induction hypothesis we know that it takes $2n-1$ days to construct the graph $K_{2n}$.Now let's take the vertices $v_i,v_j$ to show for $K_{2n+2}$. In this case we need $4n+1$ new edges.Also $2n$ vertices other than $v_i$ and $v_j $cannot match each other again.In this case, the next day let's paint $v_i $white and the remaining $ 2n+1$ vertices black and we will get $2n+1$ edges. Since $ v_j $does not create an edge with the vertices painted black the day before, let's throw $v_i $into the idle set the next day and paint $v_j$ black and paint the remaining $2n$ vertices white. In this case, we get $2n$ edge and $2$ days later we build $4n+1$ edge. So it takes $2n-1+2=2n+1$ days to construct $K_{2n+2}$Then, For $100$ vertices answer is $99$

posting for storage