Three. (25 points) From the natural numbers $1, 2, \cdots, 2010$, take $n$ numbers such that the sum of any three of the taken numbers is divisible by 21. Find the maximum value of $n$.


$$
\begin{array}{l}
\Rightarrow\left(x^{2}+16+\frac{64}{x^{2}}\right)-\left(8 x+\frac{64}{x}\right)+\frac{119}{9}=0 \\
\Rightarrow\left(x+\frac{8}{x}\right)^{2}-8\left(x+\frac{8}{x}\right)+\frac{119}{9}=0 \\
\Rightarrow\left(x+\frac{8}{x}-\frac{17}{3}\right)\left(x+\frac{8}{x}-\frac{7}{3}\right)=0 \\
\Rightarrow x=\frac{8}{3} \text { or } 3 .
\end{array}
$$

Therefore, the length of $AE$ is $\frac{8}{3}$ or 3.
Three, let $a, b, c, d$ be any four numbers taken. By the problem, we have
$$
a+b+c=21 m, a+b+d=21 n,
$$

where $m, n$ are positive integers.
Thus, $c-d=21(m-n)$.
The above equation shows that the difference between any two of the numbers taken is a multiple of 21, meaning that the remainder when each of the numbers is divided by 21 is the same.
Let this remainder be $k$. Then,
$$
\begin{array}{l}
a=21 a_{1}+k, b=21 b_{1}+k, \\
c=21 c_{1}+k,
\end{array}
$$

where $a_{1}, b_{1}, c_{1}$ are integers, $0 \leqslant k<21$.
Then $a+b+c=21\left(a_{1}+b_{1}+c_{1}\right)+3 k$.
Since $a+b+c$ is divisible by 21, $3 k$ must be divisible by 21, which means $k$ must be divisible by 7.
Therefore, $k=0, 7$ or 14.
When $k=0$, we can take $21, 42, 63, \cdots, 1995$, a total of 95 numbers, which meets the requirement;

When $k=7$, we can take $7, 28, 49, \cdots, 2002$, a total of 96 numbers, which meets the requirement;

When $k=14$, we can take $14, 35, 56, \cdots, 2009$, a total of 96 numbers, which meets the requirement.
In summary, the maximum value of $n$ is 96.