Three. (25 points) Given positive integers $a, b, c, d$ satisfying $2009(a+2009 b) c=2011(a+2011 b) d$, $(2009, d)=1,(2011, c)=1,(c, d)=1$. When $a+b$ takes the minimum value, find the values of $a, b, c, d$.

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$$
\begin{array}{l}
2009|(a+2011 b) \Leftrightarrow 2009|(a+2 b) \\
\Leftrightarrow 2009 \text { divides } 1005 a+2010 b) \\
\Leftrightarrow 2009 \text { divides } 1005 a+b) .
\end{array}
$$

Similarly,
$$
\begin{array}{l}
20111(a+2009 b) \Leftrightarrow 2011 \mid(a-2 b) \\
\Leftrightarrow 2011 \text { divides } (1005 a-2010 b) \\
\Leftrightarrow 2011 \text { divides } (1005 a+b) .
\end{array}
$$

Therefore, $(2009 \times 2011) \mid(1005 a+b)$.
Let $1005 a+b=2009 \times 2011 k\left(k \in \mathbf{N}_{+}\right)$.
Notice that
$$
\begin{array}{l}
1005(a+b)=(1005 a+b)+1004 b \\
=2009 \times 2011 k+1004 b \\
=\left(2010^{2}-1005\right) k+1004(k+b),
\end{array}
$$

and $(1005,1004)=1$, thus,
$$
1005 \text { divides } (k+b) \Rightarrow k+b \geqslant 1005 \text {. }
$$

Therefore, $1005(a+b)$
$$
\geqslant\left(2010^{2}-1005\right)+1004 \times 1005 \text {, }
$$

which means $a+b \geqslant 4020-1+1004=5023$.
When $k=1, b=1004$, the equality holds.
It is easy to verify that $a=4019, b=1004$ satisfies the given conditions.
Therefore, $a+b=5023$.
Thus, the minimum value of $a+b$ is 5023.
Substituting $a=4019, b=1004$ into the given equation yields $1005 c=1007 d$.
Since $(c, d)=1$, then $c=1007, d=1005$.
Hence, $a=4019, b=1004, c=1007, d=1005$.