7. (10 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". Among the natural numbers greater than 2000, the smallest "Eight Immortals Number" is
2016
【Solution】Solution: According to the problem, we know:
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