Example 5 Find the smallest positive integer $n$, such that there exist $n$ distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}$ simultaneously satisfying the following 3 conditions:
(1) The least common multiple of $a_{1}, a_{2}, \cdots, a_{n}$ is 2002;
(2) For any $1 \leqslant i<j \leqslant n, a_{i}$ and $a_{j}$ have a greatest common divisor greater than 1;
(3) The product $a_{1} a_{2} \cdots a_{n}$ is a perfect square and a multiple of 32. Furthermore, for this smallest positive integer $n$, find all $n$-tuples $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ that satisfy conditions (1), (2), and (3).
Since $2002=2 \times 7 \times 11 \times 13$, and $a_{i}$ is a divisor of 2002 $(i=1,2, \cdots, n)$, we have
$$
a_{i}=2^{\alpha_{i}} 7^{\beta_{i}} 11^{\gamma_{i}} 13^{\delta_{i}}(i=1,2, \cdots, n),
$$

where $\alpha_{i}, \beta_{i}, \gamma_{i}, \delta_{i} \in\{0,1\}$.
Let $a_{1} a_{2} \cdots a_{n}=k^{2}$. Since $a_{i} \mid 2002$, each $a_{i}$ is not a multiple of $2^{2}$, but $32=2^{5} \mid k^{2}$, hence $2^{6} \mid k^{2}$, so $n \geqslant 6$, and the multiples of 2 of the form (1) are only the following 8:
$$
2,2 \times 7,2 \times 11,2 \times 13,2 \times 7 \times 11,2 \times 7 \times 13,2 \times 11 \times 13,2 \times 7 \times 11 \times 13.
$$

If $n=6$, then because $a_{1}, a_{2}, \cdots, a_{6}$ are all numbers in (2) and the product of the 8 numbers in (2) is $2^{8} \times 7^{4} \times 11^{4} \times 13^{4}$, which is a perfect square, but the product of any two numbers in (2) is not a perfect square, hence the product of any 6 numbers in (2) is not a perfect square, which contradicts the given condition (3), so $n \geqslant 7$.

Assume $n=7$, then $a_{1}, a_{2}, \cdots, a_{7}$ cannot all be multiples of 2 (otherwise $2^{7} \mid k^{2}$, and $2^{8} \mid k^{2}$, a contradiction), but $2^{6} \mid k^{2}$, so among $a_{1}, a_{2}, \cdots, a_{7}$, exactly 6 are multiples of 2, i.e., exactly one of $a_{1}, a_{2}, \cdots, a_{7}$ is not in (2). Without loss of generality, let $a_{1}$ not be in (2), by the given conditions (1) and (3), $a_{1}, a_{2}, \cdots, a_{7}$ must include at least two multiples of 7, at least two multiples of 11, and at least two multiples of 13, combined with condition (2), the greatest common divisor of any two numbers in $a_{1}, a_{2}, \cdots, a_{7}$ is greater than 1. Since $a_{1}$ is not a multiple of 2, $a_{1}$ must be divisible by $7 \times 11 \times 13$. Thus, from (1), $a_{1}=7 \times 11 \times 13$, and the product of all numbers in (2) is $2^{8} \times 7^{4} \times 11^{4} \times 13^{4}$, so the only possibility is $a_{1} a_{2} \cdots a_{7}=2^{6} \times 7^{4} \times 11^{4} \times 13^{4}$, hence $a_{2} a_{3} \cdots a_{7}=2^{6} \times 7^{3} \times 11^{3} \times 13^{3}$, i.e., only two numbers in (2): 2 and $2 \times 7 \times 11 \times 13$ do not belong to $a_{2}, a_{3}, \cdots, a_{7}$, i.e., $\left\{a_{2}, a_{3}, \cdots, a_{7}\right\}=\{2 \times 7,2 \times 11,2 \times 13,2 \times 7 \times 11,2 \times 7 \times 13,2 \times 11 \times 13\}$.

In summary, the minimum value of $n$ is 7, at this time $\left\{a_{1}, a_{2}, \cdots, a_{7}\right\}=\{7 \times 11 \times 13,2 \times 7,2 \times 11, 2 \times 13,2 \times 7 \times 11,2 \times 7 \times 13,2 \times 11 \times 13\}$.