Let $p < 2008$ be a fixed prime number. There exists $r = r(p)$ such that $k^2 + k \neq r \pmod{p}$ for any integer $k$; this follows, for example, from the fact that if $k \equiv 0 \pmod{p-1}$ (mod $p$), then $k^2 + k \equiv 0 \pmod{p}$.

Now if $\{p_1, p_2, \dots, p_m\}$ is the set of all prime numbers not exceeding $2008$, we take $n$ satisfying
$$n \equiv p_j - r(p_j) \pmod{p_j}, \quad j = 1, 2, \dots, m,$$
which exists due to the Chinese Remainder Theorem. This number has the desired property.