Example 8 Given $A \subseteq\{1,2, \cdots, 2000\}$, and the absolute difference between any two numbers in $A$ is not equal to 4 or 7, find the maximum value of $|A|$.
Notice that $2000=11 \times 181+9$, we take $A=\{x \mid x=11 t+1,11 t+4$, $11 t+6,11 t+7,11 t+9,0 \leqslant t \leqslant 181, t \in \mathbf{Z}\}$, then the difference between any two numbers in $A$ is not 4 or 7. In fact, if $|11(t-r)+(b-a)|=4$ or 7, where $a, b \in\{1,4,6,7,9\}$, and $0 \leqslant r \leqslant t \leqslant 181$, then when $t-r \geqslant 2$, it is obviously not possible; when $t-r=1$, there should be $|11+(b-a)|=4$ or 7, removing the absolute value, there should be $|b-a|=4$ or 7, which can be directly verified as not possible; when $t-r=0$, similarly, it is also not possible. Therefore, the maximum value of $|A|$ is $\geqslant 5 \times 182=910$. On the other hand, suppose $A$ is a set that satisfies the condition and $|A|>910$, at this time there exists $t$, such that the set $\{x \mid x=11 t+r, 1 \leqslant r \leqslant 11\}$ contains 6 numbers, all of which, when reduced by $11 t$, belong to $A$, indicating that 1, 2, ..., 11 contain 6 numbers that are in $A$. We prove that this is impossible. Arrange 1, 2, ..., 11 in a circle, for example, 1, 5, 9, 2, 6, 10, 3, 7, 11, 4, 8 (here 1 and 8 are connected at the ends), then the absolute value of the difference between any two adjacent numbers on this circle is 4 or 7. Selecting any 6 numbers from it, there must be two numbers that are adjacent, thus 1, 2, ..., 11 cannot have 6 numbers simultaneously in $A$, leading to a contradiction. In summary, the maximum value of $|A|$ is 910.