Three. (25 points) Let $[x]$ denote the greatest integer not exceeding the real number $x$. Find the smallest positive integer $n$ such that
$$
\begin{array}{l}
{\left[\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4 n^{2}+4 n+1}-\sqrt{4 n^{2}+4 n}}\right]} \\
=[\sqrt{4 n+2018}] .
\end{array}
$$
$$
\begin{array}{l}
\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4 n^{2}+4 n+1}-\sqrt{4 n^{2}+4 n}} \\
=\frac{\sqrt{n+1}-\sqrt{n}}{2 n+1-2 \sqrt{n^{2}+n}}=\frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1}-\sqrt{n})^{2}} \\
=\sqrt{n+1}+\sqrt{n} .
\end{array}
$$

Let $k=[\sqrt{n}+\sqrt{n+1}]$. Then
$$
\begin{array}{l}
k^{2} \leqslant(\sqrt{n}+\sqrt{n+1})^{2}<4 n+2 \\
\Rightarrow k^{2} \leqslant 4 n+1 . \\
\text { By }[\sqrt{4 n+2018}]=k \\
\Rightarrow 4 n+2018<(k+1)^{2} \\
\Rightarrow 4 n+2018 \leqslant k^{2}+2 k .
\end{array}
$$

Combining $k^{2} \leqslant 4 n+1$ we get
$$
\begin{array}{l}
k^{2}+2017 \leqslant 4 n+2018 \leqslant k^{2}+2 k \\
\Rightarrow k \geqslant 1008.5 \Rightarrow 4 n+1 \geqslant 1009^{2} \\
\Rightarrow n \geqslant 1018081 .
\end{array}
$$

Therefore, the smallest positive integer $n$ is 1018081.