Let $a_{0}=12008, a_{n}=120 \underbrace{3 \cdots 3}_{n \text { 3's}} 08, n=1,2, \cdots$.
First, since
$$a_{0}=19 \times 632,$$

it follows that
$$19 \mid a_{0} .$$

Second, suppose $19 \mid a_{n}$, then from
$$\begin{array}{c} a_{n+1}-10 a_{n}=228=19 \times 12 \\ 19 \mid a_{n+1} \end{array}$$

Therefore, for all integers $n$, the number $a_{n}$ is a multiple of 19.