Three, (50 points) Find the smallest positive integer $m$, such that for any positive integer $n \geqslant m$, there always exists a positive integer $l$, such that $n<l^{2}<$
$$
\left(1+\frac{1}{2009}\right) n \text {. }
$$
Three, Prove the general conclusion. Lemma: For any positive integer $r$, when $n \geqslant m$, there exists a positive integer $l$ such that $nn$; also:
$l^{2} \leqslant(\sqrt{n}+1)^{2}=n+2 \sqrt{n}+1=n\left(1+2 \sqrt{\frac{1}{n}}+\frac{1}{n}\right)$
$\leqslant n\left(1+\frac{2}{2 r+1}+\frac{1}{(2 r+1)^{2}}\right)<n\left(1+\frac{2}{2 r+1}+\frac{1}{2 r+1} \cdot \frac{1}{r}\right)=\left(1+\frac{1}{r}\right) n$,
$\therefore \quad l=[\sqrt{n}]+1$ satisfies the condition.
In summary, $m_{\min }=4 r^{2}+1$.
Back to the problem: Take $r=2009$, we get $m_{\min }=4 \times 2009^{2}+1=16144325$.