15. Let $S=\left\{m \mid m \in \mathbf{N}_{+}, m\right.$ has prime factors no greater than 10$\}$, find the smallest positive integer $n$, such that any $n$-element subset of $S$ always contains 4 distinct numbers whose product is a perfect square.
15. Let $M$ be any $n$-element subset of $S$. According to the problem, the product of any two numbers $a$ and $b$ ($a \neq b$) in $M$ can be written as $ab = k_{ab}^2 \cdot t_{ab}$, where $k_{ab}$ is a positive integer and $t_{ab} = 2^{\alpha_1} 3^{\alpha_2} 5^{\alpha_3} 7^{\alpha_4}$ ($\alpha_i$ is 0 or 1, $1 \leq i \leq 4$). We associate $ab$ with $t_{ab}$. Since there are only $2^4 = 16$ different $t_{ab}$, and $M$ can form $C_n^2$ different pairs of elements, when $C_n^2 \geq 17$, there must exist two different pairs of numbers $(a, b)$ and $(c, d)$ in $M$ such that their corresponding $t_{ab}$ and $t_{cd}$ are equal. Thus, $abcd = k_{ab}^2 t_{ab} \cdot k_{cd}^2 \cdot t_{cd} = (k_{ab} k_{cd} t_{ab})^2$ is a perfect square. If $a, b, c, d$ are all distinct, then there exist 4 different numbers in $M$ whose product is a perfect square. Otherwise, one number in $(a, b)$ is the same as one number in $(c, d)$ while the other is different. Without loss of generality, assume $a = c$ and $b \neq d$. Then, since $abcd = a^2 bd$ is a perfect square, $bd$ is a perfect square. Consider the set $M \setminus \{b, d\}$. Similarly, when $C_{n-2}^2 \geq 17$, i.e., $n \geq 9$, either there are 4 different numbers $a_1, b_1, c_1, d_1$ in $M \setminus \{b, d\}$ whose product is a perfect square, or there are 2 different numbers $b_1, d_1$ whose product $b_1 d_1$ is a perfect square. In the latter case, there are still 4 different numbers $b, d, b_1, d_1$ in $M$ whose product is a perfect square. Therefore, when $n \geq 9$, any $n$-element subset of $S$ must contain 4 different numbers whose product is a perfect square. On the other hand, the following 8-element subset of $S$: $\{1, 2, 3, 5, 7, 2 \times 3 \times 5 \times 7, 2^3 \times 3 \times 5 \times 7, 2 \times 3^3 \times 5 \times 7\}$ does not have any 4 different numbers whose product is a perfect square. In summary, the smallest positive integer is 9.