10. Question: Among $1,2,3, \cdots, 1999,2000,2001$, what is the maximum number of numbers that can be chosen such that the sum of any three chosen numbers is divisible by 21?
10. Take at most 95.

Let $a, b, c, d$ be 4 of the numbers taken. According to the problem, $a+b+c$ and $b+c+d$ are both multiples of 21. Therefore, the difference $a - d$ is also a multiple of 21. By the arbitrariness of $a$ and $d$, it can be concluded that the difference between any two of the numbers taken is a multiple of 21.
Assume the form of the numbers taken is
$$
\begin{array}{l}
d, d+21, d+42, \cdots, d+21(n-1), d+21 n, \\
d, n \in \mathbf{Z} .
\end{array}
$$

First, the first three terms of this sequence satisfy the condition: $211(3 d+63)$. That is, 7 divides $d$. To make the sequence as long as possible, $d$ should take the smallest value, so $d=7$.
Also, $7+21 n \leqslant 2001$, so the maximum value of $n$ is 94.
Therefore, the sequence of numbers taken is $7, 28, 49, \cdots, 1960, 1981$, a total of 95 numbers.