Let us prove by induction that Alice can guarantee the existence of directed path of length $k$ after her $k$-th move. The case $k = 1$ is obvious. Suppose that prior to Alice's $k+1$-th move there is a directed path $\overrightarrow{v_0, v_1}, \overrightarrow{v_1, v_2}, \dots, \overrightarrow{v_{k-1}, v_k}$ of length $k$ and there is no directed path of length $k+1$. If Alice can not construct a cycle at her $k+1$-th move it follows that all edges connecting $v_0, \dots, v_k$ are already directed and there is a vertex $v'$ in the complement of $v_0, \dots, v_k$. Consider the edge $v', v_0$, if it is not directed Alice directs it as $v', v_0$ and enlarges the path. Consider the edge $(v', v_k)$; if it is not directed Alice directs it as $v_k, v'$ and enlarges the path. Otherwise $v', v_0$ is directed as $\overrightarrow{v_0, v'}$ and $v', v_k$ is directed as $\overrightarrow{v', v_k}$. All edges $v', v_k$, $1 \le i \le k$ can not be directed, otherwise there exists a minimal index (well defined) $l, 1 < l < k$ with $v_{l-1}, v'$ and $v', v_l$ and there is a longer directed path $v_0, v_1, \dots, v_{l-1}, v', v_l, \dots, v_k$ of length $k+1$, a contradiction. Now Alice can take an undirected path $(v', v_l)$ with minimal index $1 < l < k$ and direct it as $\overrightarrow{v', v_l}$ and get a path of length $k+1: v_0, v_1, \dots, v_{l-1}, v', v_l, \dots, v_k$. Done.

Suppose that Alice has no winning strategy, then Alice makes at least 2016 moves since $\frac{(2014)^2}{1001} > 2005$ and since prior to Alice's $t+1$-th move there is a directed path of length $t$, in order to avoid a cycle all edges connecting the vertices of this path should be directed by Bob: $\binom{t}{2} - t \le 1000t$. Thus Alice should move at most $t \le 2003$ times. Contradiction.