Do not. Let the numbers on the cards be $n_{1}, \ldots, n_{2007}$ where $n_{1} \neq n_{2}$. Let

$$s_{i}=n_{1}+n_{2}+\ldots+n_{i},$$

for $i=1,2, \ldots, 2007$. We now know that $s_{i} \equiv \equiv(\bmod 2008)$ for all $i$. Suppose that $s_{i} \equiv s_{j}$ $(\bmod 2008)$ with $i1$ means that the non-empty sum $s_{i}-n_{2}=n_{1}+n_{3}+n_{4}+\ldots+n_{i}$ is equal to 0 modulo 2008; again a contradiction.