【Answer】Solution: According to the problem:
Among the numbers $1-9$, the Eight Immortals number must be a multiple of 4 and 3, the smallest number greater than 2000 and a multiple of 12 is 2004 (a multiple of 1, 2, 3, 4, 6) but it does not meet the condition.
$2004+12=2016, 2016$ is $(1,2,3,4,6,7,8,9$'s multiple) which satisfies the condition.
Therefore, the answer is: 2016
8. (8 points) For a natural number $N$, if at least eight of the nine natural numbers from $1$ to $9$ can divide $N$, then $N$ is called an "Eight Immortals Number". The smallest "Eight Immortals Number" greater than 2000 is $\qquad$ .