Three, (50 points) Find the smallest positive integer $n$, such that any $n$-element set $A$ of positive integers contains 15 elements whose sum is divisible by 15.
Three, Solution: Let $A$ have 28 elements, with 14 elements each congruent to 1 and 2 modulo 15. If $A$ has 15 elements, the sum of which is divisible by 15, let there be $x (1 \leqslant x \leqslant 14)$ elements congruent to 1 modulo 15, and $15-x$ elements congruent to 2 modulo 15, then $x + 2(15 - x) \equiv 0 (\bmod 15)$, i.e., $x \equiv 0 (\bmod 15)$, which is a contradiction. Therefore, it must be that $n \geqslant 29$. We now prove that in any 29-element set of positive integers $A$, there are 15 elements whose sum is divisible by 15.

Lemma 1: In any 5-element set of positive integers $B$, there must be 3 elements whose sum is divisible by 3.
Proof: If $B$ has 3 elements in the same residue class modulo 3, then these 3 numbers satisfy the requirement; otherwise, $B$ contains a complete residue system modulo 3, and these 3 elements also satisfy the requirement.

Lemma 2: In any 9-element set of positive integers $C$, there must be 5 elements whose sum is divisible by 5.
Proof: Let the elements of $C$ be $x_{i} (i=1,2, \cdots, 9)$. If the sum of any 5 elements $x_{i_{1}}, x_{i_{2}}, x_{i_{3}}, x_{i_{4}}, x_{i_{5}} (1 \leqslant i_{1}<i_{2}<i_{3}<i_{4}<i_{5} \leqslant 9)$ cannot be divisible by 5, by Fermat's Little Theorem, we have
$$
\left(x_{i_{1}} + x_{i_{2}} + x_{i_{3}} + x_{i_{4}} + x_{i_{5}}\right)^{4} \equiv 1 (\bmod 5).
$$

There are $C_{9}^{5}$ such expressions. Adding them all together, we get
$$
\sum \left(x_{i_{1}} + x_{i_{2}} + x_{i_{3}} + x_{i_{4}} + x_{i_{5}}\right)^{4} \equiv C_{9}^{5} (\bmod 5).
$$
The left side of (2) is the sum of terms of the form $x_{j t}^{n} \cdots x_{j i}^{n} (1 \leqslant t \leqslant 4, n_{1}, \cdots, n_{t} \in \mathbf{N}^{*}, n_{1} + \cdots + n_{t} = 4)$, and there are $C_{9-t}^{5-t}$ expressions of the form (1) that contain $x_{j i}^{n} \cdots x_{j i}^{n}$. Since $1 \leqslant t \leqslant 4$ implies $C_{9-t}^{5-t} \equiv 0 (\bmod 5)$, the left side of (2) is congruent to 0 modulo 5, but clearly $C_{9}^{5} \equiv 1 (\bmod 5)$, which is a contradiction. Lemma 2 is proved.

Now suppose $A$ has 29 elements. By Lemma 2, we can select $x_{11}, \cdots, x_{15}$ whose sum is divisible by 5, with quotient $y_{1}$; from the remaining 24 elements, select $x_{21}, \cdots, x_{25}$ whose sum is divisible by 5, with quotient $y_{2}$; from the remaining 19 elements, select $x_{31}, \cdots, x_{35}$ whose sum is divisible by 5, with quotient $y_{3}$; from the remaining 14 elements, select $x_{41}, \cdots, x_{45}$ whose sum is divisible by 5, with quotient $y_{4}$; finally, from the remaining 9 elements, select $x_{51}, \cdots, x_{55}$ whose sum is divisible by 5, with quotient $y_{5}$.

By Lemma 1, we can select 3 elements from $y_{1}, y_{2}, y_{3}, y_{4}, y_{5}$ whose sum is divisible by 3, so the corresponding 3 groups of 15 elements in $A$ must have a sum divisible by 15.
In conclusion, the minimum value of $n$ is 29.