The nonzero numbers $x{}$ and $y{}$ satisfy the inequalities $x^{2n}-y^{2n}>x$ and $y^{2n}-x^{2n}>y$ for some natural number $n{}$. Can the product $xy$ be a negative number?

[i]Proposed by N. Agakhanov[/i]
1. We are given the inequalities \( x^{2n} - y^{2n} > x \) and \( y^{2n} - x^{2n} > y \) for some natural number \( n \). We need to determine if the product \( xy \) can be negative.
2. Suppose \( xy < 0 \). Without loss of generality, assume \( x > 0 \) and \( y < 0 \). This implies \( y = -|y| \).
3. Substitute \( y = -|y| \) into the inequalities:
   \[
   x^{2n} - (-|y|)^{2n} > x \quad \text{and} \quad (-|y|)^{2n} - x^{2n} > -|y|
   \]
4. Since \( (-|y|)^{2n} = |y|^{2n} \), the inequalities become:
   \[
   x^{2n} - |y|^{2n} > x \quad \text{and} \quad |y|^{2n} - x^{2n} > -|y|
   \]
5. Consider the first inequality \( x^{2n} - |y|^{2n} > x \):
   \[
   x^{2n} - |y|^{2n} > x
   \]
   Since \( x > 0 \), we can rewrite it as:
   \[
   x^{2n} > x + |y|^{2n}
   \]
6. Consider the second inequality \( |y|^{2n} - x^{2n} > -|y| \):
   \[
   |y|^{2n} - x^{2n} > -|y|
   \]
   Since \( |y| > 0 \), we can rewrite it as:
   \[
   |y|^{2n} > x^{2n} - |y|
   \]
7. From the first inequality, we have \( x^{2n} > x + |y|^{2n} \). This implies:
   \[
   x^{2n} - |y|^{2n} > x
   \]
   which is consistent with the given inequality.
8. From the second inequality, we have \( |y|^{2n} > x^{2n} - |y| \). This implies:
   \[
   |y|^{2n} - x^{2n} > -|y|
   \]
   which is also consistent with the given inequality.
9. However, combining both inequalities, we get:
   \[
   x^{2n} > x + |y|^{2n} \quad \text{and} \quad |y|^{2n} > x^{2n} - |y|
   \]
   This leads to a contradiction because \( x^{2n} \) cannot be simultaneously greater than \( x + |y|^{2n} \) and less than \( |y|^{2n} \).

Therefore, the assumption that \( xy < 0 \) leads to a contradiction. Hence, the product \( xy \) cannot be negative.

The final answer is \( \boxed{\text{No}} \).