Three. (25 points) It is known that the unit digit of $x_{1}^{3}+x_{2}^{3}+\cdots+x_{8}^{3}-x_{9}^{3}$ is 1, where $x_{1}, x_{2}, \cdots, x_{9}$ are nine different numbers from 2001, 2002, ..., 2009, and $8 x_{9}>$ $x_{1}+x_{2}+\cdots+x_{8}$. Find the value of $x_{9}$.

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Three, because $x_{1}, x_{2}, \cdots, x_{9}$ are nine different numbers from $2001, 2002, \cdots, 2009$, and the unit digits of $2001, 2002, \cdots, 2009$ are $1, 2, 3, 4, 5, 6, 7, 8, 9$. After cubing these unit digits, the resulting unit digits are $1, 8, 7, 4, 5, 6, 3, 2, 9$. Therefore, the unit digit of $x_{1}^{3} + x_{2}^{3} + \cdots + x_{9}^{3}$ must be 5.
Furthermore, $x_{1}^{3} + x_{2}^{3} + \cdots + x_{8}^{2} - x_{9}^{3}$
$= x_{1}^{3} + x_{2}^{3} + \cdots + x_{8}^{3} + x_{9}^{3} - 2 x_{9}^{3}$.
Assume the unit digit of $x_{9}^{3}$ is $a_{9}$.
Thus, the unit digit of $x_{1}^{3} + x_{2}^{3} + \cdots + x_{9}^{3}$ has two cases:
(1) When $5 - 2 a_{9} > 0$, it is $5 - 2 a_{9}$;
(2) When $5 - 2 a_{9} < 0$, it is $15 - 2 a_{9}$.
Since $x_{1} + x_{2} + \cdots + x_{8} < x_{1} + x_{2} + \cdots + x_{8}$, therefore, $x_{9} = 2008$.
(From the 40th Middle School of Shusheng City, Shandong Province, 277200)