= Solution
Complementing every friendship sends every degree $d$ to $n-1-d$, and hence sends $S=M+L$ to $2(n-1)-S$. Since the random graph and its complement have the same law, the distribution of $S$ is symmetric about $n-1$, giving $\mathbb ES=n-1$.
The same symmetry gives equal probabilities strictly below and above $n-1$. There is positive probability at $n-1$: for odd $n$ use a regular graph of degree $(n-1)/2$; for even $n=2m$, use $K_{m,m}$ with a perfect matching removed and then restore one matching edge. Thus the lower <median> as defined is also $n-1$.
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