You are given a positive integer $n$. What is the largest possible number of numbers that can be chosen from the set
$\{1, 2, \ldots, 2n\}$ so that there are no two chosen numbers $x > y$ for which $x - y = (x, y)$?

Here $(x, y)$ denotes the greatest common divisor of $x, y$.

[i]Proposed by Anton Trygub[/i]
1. **Claim**: The maximum number of integers that can be chosen from the set $\{1, 2, \ldots, 2n\}$ such that there are no two chosen numbers $x > y$ for which $x - y = (x, y)$ is $n$.

2. **Construction**: Consider the set of odd numbers $\{1, 3, 5, \ldots, 2n-1\}$. This set contains $n$ numbers. We need to verify that this set satisfies the condition that there are no two chosen numbers $x > y$ for which $x - y = (x, y)$.
   - For any two odd numbers $x$ and $y$ in this set, $x - y$ is even (since the difference of two odd numbers is even).
   - The greatest common divisor $(x, y)$ of two odd numbers is odd.
   - Since $x - y$ is even and $(x, y)$ is odd, it is impossible for $x - y = (x, y)$ because an even number cannot equal an odd number.
   - Therefore, the set $\{1, 3, 5, \ldots, 2n-1\}$ satisfies the condition.

3. **Proof of Maximum**: Assume for contradiction that we can choose $n+1$ numbers from the set $\{1, 2, \ldots, 2n\}$ that satisfy the condition.
   - By the pigeonhole principle, among any $n+1$ numbers chosen from $\{1, 2, \ldots, 2n\}$, there must be at least one pair of consecutive integers $c$ and $c+1$.
   - For such a pair, $(c, c+1) = 1$ because consecutive integers are coprime.
   - However, $c+1 - c = 1$, which means $x - y = (x, y)$ for $x = c+1$ and $y = c$.
   - This contradicts the condition that $x - y \neq (x, y)$.
   - Therefore, it is impossible to choose more than $n$ numbers from the set $\{1, 2, \ldots, 2n\}$ that satisfy the condition.

4. **Conclusion**: The maximum number of integers that can be chosen from the set $\{1, 2, \ldots, 2n\}$ such that there are no two chosen numbers $x > y$ for which $x - y = (x, y)$ is $n$.

The final answer is $\boxed{n}$.