(2) First, select $n$ numbers from $1,2, \cdots, 2020$, then choose any two numbers $a$ and $b$ from these $n$ numbers, such that $a \neq 2 b$. Find the maximum value of $n$.
(2) If array $A_{k}$ contains $m$ numbers, then by the Pigeonhole Principle, we know that at most $\left[\frac{m+1}{2}\right]$ numbers can be taken from $A_{k}$ such that no two numbers have a 2-fold relationship. We can calculate the maximum number of numbers that can be taken from each array and then sum them up.
For the 757 arrays from $A_{1010}$ to $A_{254}$, each group has one or two numbers. For the 190 arrays from $A_{253}$ to $A_{64}$, each group has three or four numbers. For the 47 arrays from $A_{63}$ to $A_{17}$, each group has five or six numbers. For the 12 arrays from $A_{16}$ to $A_{5}$, each group has seven or eight numbers. For the 3 arrays from $A_{4}$ to $A_{2}$, each group has nine numbers. $A_{1}$ has 11 numbers. Therefore, the maximum value of $n$ does not exceed
$$
\begin{array}{l}
757 \times 1 + 190 \times 2 + 47 \times 3 + 12 \times 4 + 3 \times 5 + 1 \times 6 \\
= 1347.
\end{array}
$$

All numbers of the form $(2 n+1) 2^{2 m} (m, n \in \mathbf{N})$ and not exceeding 2020 can be taken, totaling 1347 numbers. Among these numbers, any two numbers either have no divisibility relationship or have at least a 4-fold relationship, which meets the requirements.

In fact, these 1347 numbers can also be taken in the following way: 1010 numbers from 2020 to 1011, 253 numbers from 505 to 253, 63 numbers from 126 to 64, 16 numbers from 31 to 16, 4 numbers from 7 to 4, plus 1. Therefore, $n_{\max} = 1347$.