15. If the set $A=\{1,2, \cdots, n\}$ is arbitrarily divided into 63 pairwise disjoint subsets (which are non-empty and whose union is $A$) $A_{1}, A_{2}, \cdots, A_{63}$, then there always exist two positive integers $x$ and $y$ belonging to the same subset $A_{i}(1 \leqslant i \leqslant 63)$, such that $x>y$, and $31 x \leqslant 32 y$. Find the smallest positive integer $n$ that satisfies this condition.
15. Consider the residue classes modulo 63, i.e., partition the set \( A \) into the following 63 pairwise disjoint subsets:
$$
A_{i}=\{a \mid a=63 k+i, k \in \mathbf{N}\}(i=1,2, \cdots, 63) .
$$

Thus, for each \( A_{i}(1 \leqslant i \leqslant 63) \) and any \( x, y \in A_{i}(x>y) \), we have \( x-y \geqslant 63 \).
Then, \( y \leqslant x-63, x \leqslant n \)
$$
\begin{aligned}
\Rightarrow & 32 y-31 x \leqslant 32(x-63)-31 x \\
& =x-32 \times 63 \leqslant n-2016 .
\end{aligned}
$$

If \( n < 32 y \), it contradicts \( 31 x \leqslant 32 y \).
Hence, \( n \geqslant 32 y \) and \( x, y \) belong to the same \( A_{i} \).
Let \( x=31 \times 63 + x_{1}, y=31 \times 63 + y_{1} \),
\( 63 \geqslant x_{1} > y_{1} \geqslant 0 \).
Then \( 31 x - 32 y \)
$$
\begin{array}{l}
=31\left(31 \times 63 + x_{1}\right) - 32\left(31 \times 63 + y_{1}\right) \\
=\left(31 x_{1} - 32 y_{1}\right) + 31 \times 63(31-32) \\
\leqslant 31 x_{1} - 31 \times 63 \leqslant 31 \times 63 - 31 \times 63 = 0 .
\end{array}
$$

Thus, \( n = 2016 \) satisfies the given conditions.
In summary, the smallest value of \( n \) that satisfies the given conditions is 2016.
(Provided by Chen Deyan)