15. Find the smallest positive integer $n$ such that: if the set $A=\{1,2,3, \cdots, n\}$ is arbitrarily partitioned into 63 pairwise disjoint subsets (which are non-empty and whose union is the set $A) A_{1}, A_{2}, A_{3}, \cdots, A_{63}$, then there always exist two positive integers $x, y$ in the same subset $A_{i}(1 \leqslant i \leqslant 63)$ such that $x>y$ and $31 x \leqslant 32 y$.
Given the set $A=\{1,2,3, \cdots, n\}$, classify it using modulo 63, dividing it into 63 subsets:
$A_{i}=\{63 k+i \mid k \in \mathbf{N}\}, i=1,2,3, \cdots, 63$. Then, when $x, y \in A_{i}, x>y$, it must be that $x-y \geqslant 63 \Rightarrow y \leqslant x-63$, thus $32 y-31 x \leqslant 32(x-63)-31 x=x-32 \times 63 \leqslant n-2016$. Therefore, when $n0$, which contradicts the problem's condition, so $n \geqslant 2016$. When $n=2016$, note that among the 64 numbers $31 \times 63,31 \times 63+1,31 \times 63+2, \cdots, 31 \times 63+63$, there must be two numbers in the same subset $A_{i}$, let's say $x=31 \times 63+k, y=31 \times 63+j$, where $0 \leqslant j<k \leqslant 63$. Then $32 y-31 x=31 \times 63+32 j-31 k \geqslant 31(63-k) \geqslant 0 \Rightarrow 31 x \leqslant 32 y$. Thus, $n=2016$ satisfies the problem's condition.
In summary, the smallest positive integer $n$ that satisfies the condition is $n=2016$.