Does $\delta^+(G)+\delta^-(G) \geq n$ imply strong connectivity?
The variation that I suggested was actually slightly different variation of Woodal theorem. Perhaps I saw it in the Bang-Jensen and Gutin's book. At the time that I wrote a comment I didn't check the book for correctness. So to be sure I wrote the graph should be strongly connected. BTW, that statement holds because can be interpreted as a special case of Woodal theorem. In addition does not need strongly connectivity requirement.

This is the theorem 6.4.6 from Bang-Jensen and Gutin's book:


  Let $D$ be a digraph of order $n\ge 2$. If $\delta^+(x)+\delta^-(y) \ge n$ for all pairs of vertices $x$ and $y$ such that there is no arc
  from $x$ to $y$, then $D$ is hamiltonian.


That means the answer to the second part of your question is also Yes.

There was a doubt about that whether $n$ is a tight bound or not. Here I try to answer it. We cannot reduce the requirement of at least $n$ to $k&lt;n$, consider the following graph. $a,b,c$ are making bidirected triangle, and $e,d$ are making a bidirected $k_2$. If hamiltonian cycle starts at $e$, it cannot go to $d$ in the next move because the only way for $d$ is using $b$ but, $b$ is the only way to back to $e$. On the other hand the Hamiltonian cycle after $e$ cannot go to $c$, because then the only back way to $e,d$ is going directly to $d$ to use $b$ in next moves, but again we are in the previous position. Also from the picture it's clear that every vertex has in and out degree at least $2$. So sum of every two arbitrary in-out is at least $4=5-1 = n-1$. We can extend these sort of graphs to arbitrary $n$.



P.S1: Sure the aforementioned theorem holds for simple digraphs. i.e digraphs without loop or parallel edges.

P.S2: I don't have a good Tex tool right now. So image is not good.