For a positive integer $n$, let $1 \times 2 \times \cdots \times n=n!$ If $\frac{2017!}{2^{n}}$ is an integer, then the maximum value of $n$ is $\qquad$ .
II. 1.2 010.
From $2^{10}=1024<2017<2048=2^{11}$, we know that the number of factor 2s in 2017! is
$$
\sum_{k=1}^{10}\left[\frac{2017}{2^{k}}\right]=2010 .
$$