One, (40 points) Let $S$ be a set of positive integers with the property: for any $x \in S$, the arithmetic mean of the remaining numbers in $S$ after removing $x$ is a positive integer, and it satisfies $1 \in S$, 2016 is the maximum element of $S$. Find the maximum value of $|S|$.
Let the elements of set $S$ be
$$
1=x_{1}<x_{2}<\cdots<x_{n}=2016 \text {. }
$$

Then for $1 \leqslant j \leqslant n$, we have
$$
y_{j}=\frac{\sum_{i=1}^{n} x_{i}-x_{j}}{n-1} \in \mathbf{Z}_{+} \text {. }
$$

Thus, for $1 \leqslant i<j \leqslant n$, we have
$$
\begin{array}{l}
y_{i}-y_{j}=\frac{x_{j}-x_{i}}{n-1} \in \mathbf{Z}_{+} \\
\Rightarrow(n-1) \mid\left(x_{j}-x_{i}\right) .
\end{array}
$$

In particular, we have
$$
(n-1)|(2016-1) \Rightarrow(n-1)| 5 \times 13 \times 31 \text {. }
$$

Notice that, for any $2 \leqslant j \leqslant n$, we have
$$
(n-1) \mid\left(x_{j}-1\right) \text {. }
$$

Thus, for $x_{n}-1$ we have
$$
\begin{array}{l}
x_{n}-1=\sum_{k=1}^{n-1}\left(x_{k+1}-x_{k}\right) \geqslant(n-1)^{2} \\
\Rightarrow(n-1)^{2} \leqslant 2015 \\
\Rightarrow n \leqslant 45 . \\
\text { Also, }(n-1) \mid 2015, \text { so } n \leqslant 32 . \\
\text { Let } x_{i}=31 i-30(1 \leqslant i \leqslant 31), x_{32}=2016,
\end{array}
$$

It is easy to verify that $S$ satisfies the conditions in this case.
Therefore, $|S|_{\max }=32$.