Let $n$ be a positive integer, $[x]$ be the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$.
(1) Find all positive integers $n$ that satisfy $\sum_{k=1}^{2013}\left[\frac{k n}{2013}\right]=2013+n$;
(2) Find all positive integers $n$ that maximize $\sum_{k=1}^{2013}\left\{\frac{k n}{2013}\right\}$, and determine this maximum value.
(1) Let $m=2013, d=(m, n), m=d m_{1}, n=d n_{1}$.
For $1 \leqslant k \leqslant m-1$, we have
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]=\left[\frac{k n_{1}}{m_{1}}\right]+\left[\frac{(m-k) n_{1}}{m_{1}}\right] .
$$

When $m_{1} \mid k$
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]=n,
$$

and there are $d-1$ values of $k$ satisfying $1 \leqslant k \leqslant m-1$.
When $m_{1} \nmid k$
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]=n-1,
$$

and there are $m-1-(d-1)=m-d$ values of $k$ satisfying $1 \leqslant k \leqslant m-1$.
Thus
$$
\begin{aligned}
\sum_{i=1}^{m}\left[\frac{k n}{m}\right]= & n+\sum_{k=1}^{m-1}\left[\frac{k n}{m}\right]=n+\frac{1}{2} \sum_{k=1}^{m-1}\left(\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]\right) \\
& =n+\frac{1}{2}[n(d-1)+(n-1)(m-d)]=\frac{1}{2}(m n+n-m+d) .
\end{aligned}
$$

Therefore, when $m=2013$, we have
$$
\frac{1}{2}(2013 n+n-2013+d)=2013+n,
$$

which simplifies to $3 \times 2013=2012 n+(2013, n)>2012 n$. Thus, $n \leqslant 3$, and upon verification, $n=3$.
(2) From (1), we get
$$
\begin{aligned}
\sum_{k=1}^{m}\left\{\frac{k n}{m}\right\} & =\sum_{k=1}^{m}\left(\frac{k n}{m}-\left[\frac{k n}{m}\right]\right)=\frac{n}{m} \cdot \frac{m(m+1)}{2}-\sum_{k=1}^{m}\left[\frac{k n}{m}\right] \\
& =\frac{n(m+1)}{2}-\frac{1}{2}(m n+n-m+d)=\frac{m-d}{2} \\
& \leqslant \frac{m-1}{2} .
\end{aligned}
$$

In particular, for $m=2013$, when $d=(2013, n)=1$, i.e., when $n$ is not divisible by $3,11,61$, the maximum value of $\sum_{k=1}^{2013}\left\{\frac{k n}{2013}\right\}$ is $\frac{2013-1}{2}=1006$.