7. For a real number $x$, $[x]$ denotes the greatest integer not exceeding the real number $x$. For some integer $k$, there are exactly 2008 positive integers $n_{1}, n_{2}, \cdots, n_{2008}$, satisfying
$$
k=\left[\sqrt[3]{n_{1}}\right]=\left[\sqrt[3]{n_{2}}\right]=\cdots=\left[\sqrt[3]{n_{2008}}\right],
$$

and $k \mid n_{i}(i=1,2, \cdots, 2008)$. Then $k=$ $\qquad$
7.668.

If $\sqrt[3]{n}-1n \text {. }$

Also, if $k \mid n$, then $n$ can take $k^{3}, k^{3}+k, \cdots, k^{3}+$ $3 k^{2}+3 k$, a total of $3 k+4$.
Therefore, $3 k+4=2008 \Rightarrow k=668$.