3. 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 2015 good partitions.
3. 3024.

Let $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}\left(a_{1}<a_{2}<\cdots<a_{n}\right)$.
For any non-empty finite set of positive integers $B$, let $\operatorname{lcm} B$ and $\operatorname{gcd} B$ denote the least common multiple and greatest common divisor of the elements in $B$, respectively.

Consider any good partition $\left(A_{1}, A_{2}\right)$ of $A$. By definition, there exists a positive integer $d$ such that
$\operatorname{lcm} A_{1}=d=\operatorname{gcd} A_{2}$.
For any $a_{i} \in A_{1}$ and $a_{j} \in A_{2}$, we have
$a_{i} \leqslant d \leqslant a_{j}$.
Thus, there exists a positive integer $k(1 \leqslant k<n)$ such that
$A_{1}=\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$,
$A_{2}=\left\{a_{k+1}, a_{k+2}, \cdots, a_{n}\right\}$.
Therefore, each good partition is determined by an element $a_{k}$ $(1 \leqslant k<n)$, which we call a "separator".
Define $l_{k}=\operatorname{lcm}\left(a_{1}, a_{2}, \cdots, a_{k}\right)$,
$g_{k}=\operatorname{gcd}\left(a_{k+1}, a_{k+2}, \cdots, a_{n}\right)$,

where $1 \leqslant k \leqslant n-1$. Then $a_{k}$ is a separator if and only if $l_{k}=g_{k}$.

To prove some properties of separators, we need the following lemma.

Lemma If $a_{k-1}$ and $a_{k}$ $(2 \leqslant k \leqslant n-1)$ are both separators, then $g_{k-1}=g_{k}=a_{k}$.
Proof Assume $a_{k-1}$ and $a_{k}$ are both separators.
$$
\begin{array}{c}
\text { By } l_{k-1}=g_{k-1} \Rightarrow l_{k-1} \mid a_{k} \\
\Rightarrow g_{k}=l_{k}=\operatorname{lcm}\left(l_{k-1}, a_{k}\right)=a_{k}, \\
g_{k-1}=\operatorname{gcd}\left(a_{k}, g_{k}\right)=a_{k} .
\end{array}
$$

The lemma is proved.
Property 1 For each $k=2,3, \cdots, n-2$, at least one of $a_{k-1}$, $a_{k}$, and $a_{k+1}$ is not a separator.
Proof of Property 1 By contradiction.
Assume $a_{k-1}$, $a_{k}$, and $a_{k+1}$ are all separators.
By the lemma, we have $a_{k+1}=g_{k}=a_{k}$, which is a contradiction.
Property 2 $a_{1}$ and $a_{2}$ cannot both be separators, and $a_{n-2}$ and $a_{n-1}$ cannot both be separators.
Proof of Property 2 Assume $a_{1}$ and $a_{2}$ are both separators.
By the lemma, we have $a_{2}=g_{1}=l_{1}=\operatorname{lcm}\left(a_{1}\right)=a_{1}$, which is a contradiction.

Similarly, assume $a_{n-2}$ and $a_{n-1}$ are both separators. By the lemma, we have $a_{n-1}=g_{n-1}=\operatorname{gcd}\left(a_{n}\right)=a_{n}$, which is a contradiction.
Properties 1 and 2 are proved.
Suppose a set $A$ with $n$ elements has exactly 2015 good partitions. Clearly, $n \geqslant 5$.

By Property 2, at most one element in $\left\{a_{1}, a_{2}\right\}$ and $\left\{a_{n-2}, a_{n-1}\right\}$ can be a separator.

Let $\lfloor x\rfloor$ denote the greatest integer not exceeding the real number $x$, and $\lceil x\rceil$ denote the smallest integer not less than the real number $x$.

By Property 1, at least $\left\lfloor\frac{n-5}{3}\right\rfloor$ elements in $\left\{a_{3}, a_{4}, \cdots, a_{n-3}\right\}$ are not separators. Therefore, $A$ has at most
$$
(n-1)-2-\left\lfloor\frac{n-5}{3}\right\rfloor=\left\lceil\frac{2(n-2)}{3}\right\rceil
$$

separators.
$$
\text { Therefore, }\left\lceil\frac{2(n-2)}{3}\right\rceil \geqslant 2015 \Rightarrow n \geqslant 3024 \text {. }
$$

Finally, we prove that there exists a set $A$ with 3024 elements that has exactly 2015 separators.
In fact,
$$
A=\left\{2 \times 6^{i}, 3 \times 6^{i}, 6^{i+1} \mid 10 \leqslant i \leqslant 1007\right\}
$$

satisfies $|A|=3024$, and $3 \times 6^{i} (0 \leqslant i \leqslant 1007)$ and $6^{i} (1 \leqslant i \leqslant 1007)$ are all separators.
In summary, the minimum value of $n$ is 3024.