Let in a cell with number $n$ number $a_n$ is written down. We will consider such sequence $(a_n)$: $a_{2k-1} = 2k-1$, $k = \overline{1,1004}$, $a_{2k} = 2008 - 2k$, $k = \overline{1,1003}$. We will prove that such placing approaches us. Let $S_n = a_1 + a_2 + \dots + a_n$. Then $S_{2k} = 2007k$, $k = \overline{1,1003}$, $S_{2k-1} = S_{2k-2} + a_{2k-1} = 2009k - 2008$, $k = \overline{1,1004}$. And as $S_{2k-1} \equiv -2007k \pmod{2008}$, from equality $S_n \equiv S_M \pmod{2008}$ follows that $m = n$. Therefore the sequence $(a_n)$ satisfies the condition that for any non-negative integer numbers $i$ and $j$ such that $1 \le i < j \le 2007$, the sum of numbers of cells $i, i+1, \dots, j$ does not divide by $2008$.

Now we will describe a winning strategy for the first player. The first move he marks a cell with number $1004$. Further, if the second player moves in the cell with number $i$, the first player moves into the cell with number $2008-i$. As the sum of numbers with numbers $i$ and $2008-i$ is equal to $2008$ for all $i = \overline{1,2007}$, it is easy to see that the described strategy for the first player is winning.

We will show how from this sequence to receive $999$ more other sequences which also satisfy the statement of the problem. Let $(l, 2008) = 1$. We put $b_n = l \cdot a_n \pmod{2008}$, $n = \overline{1,2007}$. Then the sequence $(b_n)$ obviously satisfies the condition that $\forall i, j \in \mathbb{N}: 1 \le i < j \le 2007$, the sum of numbers of cells $i, i+1, \dots, j$ does not divide by $2008$. On the other hand, the strategy of the first player does not change. It is obvious that for different $l = \overline{1,2007}$, we receive different sequences $(b_n)$. Therefore, we receive $\varphi(2008) = 1000$ sequences.