The answer is $671$.

First consider a network that has a single cycle $\{C_1, C_2, C_3, C_4, C_5, C_6\}$, and has chains of $668$, $667$ and $667$ computers starting at $C_1$, $C_3$ and $C_5$, respectively. Hacker takes $C_1$ in the first move. Then if the administrator takes $C_4$, the moves $C_2$, $C_3$, $C_6$, $C_5$ guarantee the hacker $671$ computers. Hacker does better for any other response by the administrator. For instance, if the administrator takes $C_2$ in the second move, then the moves $C_6$, $C_3$, $C_5$, $C_4$ give the hacker $1338$ computers.

Now consider an arbitrary network with $2008$ computers and without intersecting cycles.

Case 1: There is a computer that is not on a cycle such that, when it is removed from the network, each of the remaining connected components contains at most $1337$ computers. Then the hacker guarantees to hack into at least $2008 - 1337 = 671$ computers by hacking into this computer in the first move.

Case 2: There is a cycle $Z = \{C_1, C_2, \dots, C_m\}$ such that, when it is removed from the network, each of the remaining connected components contains at most $1337$ computers. For $1 \le i \le m$, let $H_i$, respectively $G_i$, be the set of all hacked computers in $Z$, respectively in the entire network, when the hacker takes $C_i$ in the first move and from there on both follow their best strategies. Then $|H_i| = \lfloor m/2 \rfloor$. Take $i, j$ such that $|H_i \cup H_j|$ is maximum. If $Z = H_i \cup H_j$, then $|G_i| + |G_j| \ge 2008$, and one of the sets $G_i, G_j$ has at least $1004$ elements. If, on the other hand, $C_k \in Z \setminus (H_i \cup H_j)$, then $H_i \cup H_j \cup H_k = Z$. In this case, $|G_i| + |G_j| + |G_k| \ge 2008+3$, and therefore, one of the sets $G_i, G_j, G_k$ has at least $\lfloor 2011/3 \rfloor = 671$ elements. In either case the hacker has a strategy guaranteeing at least $671$ hacked computers.

Finally, either *Case 1* or *Case 2* must hold. Otherwise, we can move along the network by moving into the component with more than $1337$ computers at each step. At some step we must reverse our direction. When this happens we have more than $1337$ computers to each side of the connection along which we retraced our last step, a contradiction.