Three, (25 points) Given that $3 n^{3}+2013(n>1)$ is divisible by 2016. Find the smallest positive integer $n$.

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Translation:

Three, (25 points) Given that $3 n^{3}+2013(n>1)$ is divisible by 2016. Find the smallest positive integer $n$.
From the given, we have $2016 I \left(3 n^{3}-3\right)$.
Then $672 \mid (n-1)\left(n^{2}+n+1\right)$,
$$
672=2^{5} \times 3 \times 7 \text {. }
$$

Since $2 \nmid\left(n^{2}+n+1\right)$, we know $n=32 c+1$.
Because $31\left(n^{3}-1\right)$, so,
$$
\begin{array}{l}
n^{3}-1 \equiv(32 c+1)^{3}-1 \equiv-c^{3} \equiv 0(\bmod 3) \\
\Rightarrow c=3 d, n=96 d+1 . \\
\text { Also } 7 \text { I }\left(n^{3}-1\right) \text {, then } \\
n^{3}-1 \equiv(-2 d+1)^{3}-1 \equiv 0(\bmod 7) .
\end{array}
$$

Since $7 \mid\left(n^{3}-1\right)$, then

When $d=1$, it does not satisfy equation (1);
When $d=2$, equation (1) holds.
Therefore, the minimum value of $n$ is 193.