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.
I will rephrase your question slightly. Let $K_{n}^{*}$ be the directed graph with $n$ vertices and two oppositely directed edges for each pair of vertices. Your question is then the following.
What is the maximum number of edge-disjoint directed Hamiltonian cycles of $K_{n}^{*}$?
For $n=2k+1$ odd, it is an old theorem of Walecki that $K_n$ can be decomposed into $k$ Hamiltonian cycles, and hence $K_n^*$ can be decomposed into $2k$ directed Hamiltonian cycles.
For $n=2k$ even, you are right to note that for $n=4$ we cannot achieve the upper bound of $n-1.$ One can also check that we cannot achieve the upper bound for $n=6$. However, Tilson proved that for even $n \geq 8$, $K_n^*$ can de decomposed into $n-1$ directed Hamiltonian cycles.
This completely answers your question. Namely, $n=4$ and $n=6$ are the only exceptions.
Best Answer
Not always. I claim that for large $n$ a random tournament on $n$ vertices satisfies your property with probability tending to 0.
We use the following
Lemma. Consider a random tournament on $m$ vertices $1,\ldots,m$. For a given sequence $(d_1,\ldots,d_m)$ the probability that $\deg(i)=d_i$ for all $i=1,2,\ldots,m$ is at most $\exp(-cm\log m)$ (for a universal constant $c$).
Proof. Fix what happens between the first $m_1:=\lceil m/2\rceil$ vertices. Then the event that these $m_1$ vertices have prescribed degrees is an intersection of $m_1$ independent events (corresponding to these vertices) each of which has probability at most $O(m^{-1/2})$ (by de Moivre — Laplace local central limit theorem: each event for a fixed vertex is that the sum of $m-m_1$ independent 0-1's has prescribed value).
Return to your problem. The probability that a given set $S$ satisfies your condition for all $v\notin S$ is at most $(1/2)^{n-|S|}$. The sum over all $S$ of size, say, less than $n/10$, is clearly $o(1)$. For $|S|\geqslant n/10$ we look at vertices $v\in S$, that is, we are interested only in a random tournament induced on $S$. I claim that the probability that it satisfies your requirement is $o(1/2^n)$, that suffices for my initial claim.
Denote $m=|S|$. Your requirement yields that the indegree of each vertex $v\in S$ is at least $m/2-1$. If $m$ is odd, this means that the indegree is at least $(m-1)/2$, thus all indegrees are equal to $(m-1)/2$ (since it is the average indegree), and we just apply Lemma. If $m=2m_1$ is even, then all indegrees must be at least $m_1-1$. Enumerate $S$ and denote the indegrees $m_1-1+\varepsilon_i$ ($i=1,2,\ldots,2m_1$). Then $\sum \varepsilon_i=m_1$. Such a sequence of non-negative $\varepsilon_i$'s may be chosen by ${3m_1-1\choose 2m_1-1}$ ways, which is only exponential in $m$, while the Lemma says that for a fixed degree sequence the probability that it arises in a random tournament is superexponentially small.