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.)
The maximum number of edges in a strongly-connected digraph with $n \geq k+1$ vertices and no cycles of length at most $k$ is $${\binom{n}{2}} - n(k-2) + \frac{(k+1)(k-2)}{2}.$$
(A digraph where every vertex is reachable from every other vertex by a directed path is called strongly connected.)
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.
A quick check with Mathematica seems to suggest that a (smallest) counterexample is given by a graph with 6 vertices and 13 edges:
Here the red edges are contained in the most odd cycles (16, 12, 6, 5, 2, 2, 1, 0 respectively) and one possible sequence of edge removals is shown. However, choosing different edges one can end up with a bipartite graph with 7 edges.
Edit: Any sequence of edge removals of the following graph with 8 vertices and 18 edges seems to result in a bipartite graph with at most 8 edges:
Edit: For those interested: here's the Mathematica code to check that the last graph is really a counterexample.
Edit: The first complete graph for which the algorithm fails is the complete graph on 10 vertices. Indeed, it seems that the largest bipartite graph one can obtain from an $n$-vertex ($n>4$) complete graph is the complete bipartite graph on $(n-3)+3$ vertices, which has $3(n-3)$ edges, i.e. smaller than $n(n-1)/4$ for $n\geq 10$.
Best Answer
There are $$ \frac{\prod_{i=0}^{k-1}(n-i)^2}{k} $$ possible directed cycles with $2k$ vertices. Each such cycle occurs with probability $n^{-2k}$, so the exact expectation of the number of cycles is $$ \sum_{k=1}^n \frac{\prod_{i=0}^{k-1}(n-i)^2}{kn^{2k}} = \sum_{k=1}^n \frac{\prod_{i=0}^{k-1}(1-i/n)^2}{k}. $$ As $n\to\infty$, that sum is asymptotic to the integral $$ \int_1^\infty \exp(-x^2/n)/x~dx = \left[ -\frac12 Ei(1,x^2/n) \right]_1^\infty = \frac12\log n + O(1), $$ where $Ei$ is the exponential integral function and I'm relying on Maple a bit.