Three. (50 points) For a finite set $A$ consisting of positive integers, if $A$ is divided into two non-empty disjoint subsets $A_{1}$ and $A_{2}$, and the least common multiple (LCM) of the elements in $A_{1}$ equals the greatest common divisor (GCD) of the elements in $A_{2}$, then such a partition is called "good." Find the minimum value of the positive integer $n$ such that there exists a set of $n$ positive integers with exactly 2021 good partitions.
Three, obviously, because the least common multiple of set $A_{1}$ is not less than any number in $A_{1}$, and the greatest common divisor of set $A_{2}$ is not greater than any number in $A_{2}$, so any number in $A_{1}$ is less than any number in $A_{2}$.
$$
\begin{array}{l}
\text { Let } A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}, \\
A_{1}=\left\{a_{1}, a_{2}, \cdots, a_{x}\right\}, \\
A_{2}=\left\{a_{x+1}, a_{x+2}, \cdots, a_{n}\right\},
\end{array}
$$

where $1 \leqslant x \leqslant n$ is an integer.
Let $X$ be the set of all integers $x$ that satisfy the condition.
Below we prove that for any integer $c$, if
$$
\begin{array}{l}
c \in X, c+1 \in X, \\
\text { then } c+2 \notin X, c-1 \notin X . \\
\text { By }\left[a_{1}, a_{2}, \cdots, a_{c}\right]=a_{c}, \\
{\left[a_{1}, a_{2}, \cdots, a_{c+1}\right]=a_{c+1},}
\end{array}
$$

the conclusion is proved.
Thus, $n \geqslant 3033$.
Furthermore, when
$$
A=\left\{2 \times 6^{i}, 3 \times 6^{i}, 6^{i+1} 10 \leqslant i \leqslant 1010\right\}
$$

the requirement is satisfied, so the minimum value sought is 3033.
Therefore, the minimum value of $n$ is 3033.