14. A. From the 2010 positive integers $1,2, \cdots, 2010$, what is the maximum number of integers that can be selected such that the sum of any three selected numbers is divisible by 33?
14. A. First, the following 61 numbers: $11, 11+33, 11+2 \times 33, \cdots, 11+60 \times 33$ (i.e., 1991) satisfy the conditions of the problem.

On the other hand, let $a_{1}<a_{2}<\cdots<a_{n}$ be the numbers selected from 1, 2, $\cdots, 2010$ that satisfy the conditions of the problem. For any four numbers $a_{i}, a_{j}, a_{k}, a_{m}$ among these $n$ numbers, since
$$
33 \mid \left(a_{i}+a_{k}+a_{m}\right), 33 \mid \left(a_{j}+a_{k}+a_{m}\right),
$$
it follows that $33 \mid \left(a_{j}-a_{i}\right)$.

Therefore, the difference between any two of the selected numbers is a multiple of 33.
Let $a_{i}=a_{1}+33 d_{i} (i=1,2, \cdots, n)$.
From $33 \mid \left(a_{1}+a_{2}+a_{3}\right)$, we get
$33 \mid \left(3 a_{1}+33 d_{2}+33 d_{3}\right)$.
Thus, $33 \mid 3 a_{1}$, and $11 \mid a_{1}$, which means $a_{1} \geqslant 11$.

Therefore, $d_{i}=\frac{a_{i}-a_{1}}{33} \leqslant \frac{2010-11}{33}<61$.
Hence, $d_{i} \leqslant 60$. So, $n \leqslant 61$.
In conclusion, the maximum value of $n$ is 61.