We consider the class $C$ of directed simple (no multiple edges) graphs having the property that every vertex is reachable by a directed path from every other vertex.
Given an integer $k$, what is the maximal possible number of (directed) edges in a graph of $C$ with $n$ vertices
such that there are no directed cycles of length $\leq k$?
For $k=2$ this means simply
that the existence of an edge from $v$ to $w$ forbids the existence of an edge from $w$ to $v$ and one can thus choose arbitrary orientions (giving rise to a graph in $C$) on the edges of the complete unoriented graph.
For $k=3$, one has also to forbid oriented triangles which is not possible by orienting all edges of a complete graph on $n\geq 3$ vertices such that the result is in $C$.
On the other hand, there are of course no triangles by choosing arbitrary orientations
(giving rise to an element in $C$) of a complete bipartite graph.
There are thus such graphs having roughly $n^2/4$ directed edges.
There should be better upper and lower bounds.
Motivation: G. Higman (A finitely generated infinite simple group. J. London Math. Soc. 26,
(1951). 61–64) constructed finitely generated infinite simple groups
by considering quotients of the finitely presented group $$\langle g_1,\dots,g_n|g_{i-1}^{-1}g_ig_{i-1}=g_i^2\rangle$$
where indices are modulo $n$.
This group is trivial for $n=2,3$ and infinite for $n\geq 4$. Given a directed graph,
one can consider the corresponding group-presentation with generators corresponding to vertices and directed edges corresponding to relations $a^{-1}ba=b^2$. The triviality
of the group constructed by Higman associated to $n=2,3$ implies that we want to avoid
oriented cycles of length $\leq 3$ when searching for interesting examples.
Best Answer
Updated 4/17/11:
(Originally, this answer contained a different proof of the result below for $k=3$. Not only did the proof not generalize, but it was wrong.)
Gordon Royle conjectured this bound an gave an example achieving it for $k=3$. For general $k$ and $n$ the bound is attained by the following construction, almost identical to the one provided by Nathann Cohen in the comments:
Let vertices $x_1,x_2,\ldots,x_{n-k+2}$ form a transitive tournament with $x_i \to x_j$ being an edge for all $1 \leq i < j \leq n-k+2$. Now delete the edge $x_1 \to x_{n-k+2}$ and replace it with a path $x_{n-k+2} \to x_{n-k+3} \to \ldots \to x_n \to x_1$. (The vertices $x_{n-k+3},\ldots, x_n$ will have in-degree one and out-degree one in the resulting graph.)
It remains to prove that the above number is a valid upper bound. The proof is by induction on $n$.
Simple counting shows that the bound is valid if $G$ is a directed cycle. It is tight if $G$ is a cycle of length $k+1$. Assume now that $G$ is not a cycle. Then there exist $\emptyset \neq X \subsetneq V(G)$ such that $G|X$ is strongly connected. (For example, one can choose the vertex set of any induced cycle in $G$.) Choose $X$ maximal subject to the above. Let $u \to v_1$ be an edge of $G$ with $u \in X$, $v_1 \not \in X$, and let $P$ be a shortest path in $G$ from $v_1$ to $X$. Let $P=v_1 \to v_2 \to \ldots \to v_l \to w$.
Note that adding to $G|X$ any path starting and ending in $X$ produces a strongly connected digraph. It follows from the choice of $X$ that any non-trivial such path must include all the vertices in $V(G)-X$. In particular, if $l\geq 3$, $v_2,\ldots,v_{l-1}$ have no neighbors in $X$.
Let us further assume that $u$ and $w$ are chosen so that the directed path $Q$ from $w$ to $u$ in $G|X$ is as short as possible. (Perhaps, $w=u$.) Then $V(P) \cup V(Q)$ induces a cycle in $G$, and so $v_1$ and $v_l$ have at least $k-2$ non-neighbors on $V(P) \cup V(Q)$. At least $k-l$ of those non-neighbors are in $X$ if $l\geq 2$. Therefore there are at least $k-2$ non-edges (pairs of non-adjacent vertices) between $X$ and $V(G)-X$ if $l=1$, and at least $$2(k-l)+(l-2)(k+1) \geq l(k-2)$$ non-edges if $l \geq 2$. By the induction hypothesis there are at least $|X|(k-2)- \frac{(k+1)(k-2)}{2}$ non-edges between vertices of $X$, and therefore at least $$(l+|X|)(k-2)- \frac{(k+1)(k-2)}{2}=n(k-2) - \frac{(k+1)(k-2)}{2}$$ non-edges in total, as desired.