6. Given that $2^{n}\left(n \in \mathbf{N}_{+}\right)$ can divide $2007^{2018}-1$. Then the maximum value of $n$ is ( ).
(A) 12
(B) 13
(C) 14
(D) 15
6. C.

Let $2007=a$. Then
$$
\begin{array}{l} 
2007^{2048}-1=a^{2048}-1^{2} \\
=\left(a^{2^{10}}+1\right)\left(a^{2^{9}}+1\right) \cdots\left(a^{2^{1}}+1\right) . \\
(a+1)(a-1) .
\end{array}
$$

For any positive integer $k$, we have
$$
a^{2^{k}}+1 \equiv(-1)^{2^{k}}+1 \equiv 2(\bmod 4) \text {, }
$$

which means $a^{2^{k}}+1$ is a multiple of 2, but not a multiple of 4. And $a+1$ is a multiple of 8, but not a multiple of 16, while $a-1$ is a multiple of 2, but not a multiple of 4. Therefore, the maximum value of $n$ is
$$
10 \times 1+3+1=14 \text {. }
$$