Observe that $-2008 \equiv 6 \equiv 5^2 \pmod{19}$. Thus,
$$\begin{aligned} 2 \sum_{k=0}^{n} \binom{2n+1}{2k+1} (-2008)^k &\equiv 2 \sum_{k=0}^{n} \binom{2n+1}{2k+1} 5^{2k} \pmod{19} \\ &\equiv (1+5)^{2n+1} - (1-5)^{2n+1} \pmod{19} \\ &\equiv 6^{2n+1} + 4^{2n+1} \\ &\equiv 2^{2n+1} (3^{2n+1} + 2^{2n+1}) \pmod{19}. \end{aligned}$$
Since
$$3^{2n+1} + 2^{2n+1} \equiv (-16)^{2n+1} + 2^{2n+1} \equiv 2^{2n+1} (1 - 2^{6n+3}) \pmod{19}$$
and $2^{18} \equiv 1 \pmod{19}$. We can see that
$$2^{6(n+3)+3} \equiv 2^{6n+3} \pmod{19}$$
for each $n = 0, 1, 2, \dots$.
Therefore, it suffices to consider the divisibility of $3^{2n+1} + 2^{2n+1}$ by $19$ when $n = 0, 1, 2$. We now verify that
$$3^1 + 2^1 \equiv 5 \pmod{19}$$
$$3^3 + 2^3 \equiv 35 \equiv 16 \pmod{19}$$
$$3^5 + 2^5 \equiv 275 \equiv 9 \pmod{19}.$$
Hence, $19 \nmid \sum_{k=0}^{n} \binom{2n+1}{2k+1} (-2008)^k$.