You've already shown that every vertex $v$ has even degree, for if it had odd degree, than look at its set of neighbors with the induced subgraph structure, $H$. $H$ has an odd number of vertices with every vertex having odd degree, which is a contradiction.
Now, consider the adjacency matrix $A$ of $G,$ where we consider a vertex to be adjacent to itself.
Then the condition of $|N(u)\cap N(v)|$ being odd translates to $A^2=F,$ where $F$ is the matrix with each entry being 1. Since every vertex of $A$ has even degree, we have the identity $AF=F$. Therefore $F^2=FA^2=F$. The identity $F^2=F$ exactly means that the number of vertices is odd. This completes the proof.
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.
Best Answer
Take a Steiner triple system on $v$ points. Let $X$ be the graph with the $v(v-1)/6$ triples as its vertices, two triples adjacent if the have exactly one point in common. We need $v\equiv1,3$ modulo 6. Then two adjacent triples have exactly $(v+3)/2$ common neighbours, and two disjoint triples have exactly 9 common neighbours. If we take $v\equiv3,7$ modulo 12 we get examples.
Of course I am just constructing strongly regular graphs with $\lambda$ and $\mu$ odd. The are strongly regular graphs with this property besides the ones listed, for example generalized quadrangles with $s$ and $t$ even. Further examples appear in Andries Brouwer's on-line tables (http://www.win.tue.nl/~aeb/graphs/srg/srgtab.html), or Gordon Royle's (http://units.maths.uwa.edu.au/~gordon/remote/srgs/).
This suggest that a classification might be difficult.