Three, (25 points) In the set of numbers $1,2, \cdots, 2009$, what is the maximum number of numbers that can be selected such that the sum of any two selected numbers is divisible by 100?
Three, let the $n$ numbers that meet the conditions be,
$$
a_{1}, a_{2}, \cdots, a_{n} \text {, }
$$

Take any three of these numbers (let them be $a_{k} \backslash a_{m} \backslash a_{f}$). Then
$$
\begin{array}{l}
a_{k}+a_{m}=100 k_{1}, \\
a_{k}+a_{4}=100 k_{2}, \\
a_{m}+a_{4}=100 k_{3},
\end{array}
$$

where $k_{1} 、 k_{2} 、 k_{3}$ are positive integers.
$$
\begin{array}{l}
\text { (1) }+ \text { (2)-(3) gives } \\
a_{k}=50\left(k_{1}+k_{2}-k_{3}\right) .
\end{array}
$$

Similarly, $a_{m}=\mathbf{5 0}\left(k_{1}+k_{3}-k_{2}\right)$,
$$
a_{1}=50\left(k_{2}+k_{3}-k_{1}\right) \text {. }
$$

Therefore, $a_{k} 、 \grave{a}_{m} 、 a$ are all multiples of 50.
However, the multiples can only be odd or even, and they must all be odd or all even to be divisible by 100, so the numbers that meet the conditions are $50,150, \cdots, 1950$ or $100,200, \cdots, 2000$, with 20 numbers in each set.