Let $n$ be a natural number. Find all real numbers $x$ satisfying the equation $$\sum^n_{k=1}\frac{kx^k}{1+x^{2k}}=\frac{n(n+1)}4.$$
1. We start with the given equation:
   \[
   \sum_{k=1}^n \frac{kx^k}{1 + x^{2k}} = \frac{n(n+1)}{4}
   \]

2. Let's denote this equation as \( P(n) \):
   \[
   P(n) \implies \sum_{k=1}^n \frac{kx^k}{1 + x^{2k}} = \frac{n(n+1)}{4}
   \]

3. Consider the difference \( P(n+1) - P(n) \):
   \[
   \sum_{k=1}^{n+1} \frac{kx^k}{1 + x^{2k}} - \sum_{k=1}^n \frac{kx^k}{1 + x^{2k}} = \frac{(n+1)(n+2)}{4} - \frac{n(n+1)}{4}
   \]

4. Simplifying the right-hand side:
   \[
   \frac{(n+1)(n+2) - n(n+1)}{4} = \frac{(n+1)(n+2 - n)}{4} = \frac{(n+1) \cdot 2}{4} = \frac{n+1}{2}
   \]

5. Therefore, we have:
   \[
   \frac{(n+1)x^{n+1}}{1 + x^{2(n+1)}} = \frac{n+1}{2}
   \]

6. Simplifying the left-hand side:
   \[
   \frac{x^{n+1}}{1 + x^{2(n+1)}} = \frac{1}{2}
   \]

7. Cross-multiplying gives:
   \[
   2x^{n+1} = 1 + x^{2(n+1)}
   \]

8. Rearranging terms:
   \[
   x^{2(n+1)} - 2x^{n+1} + 1 = 0
   \]

9. This is a quadratic equation in terms of \( y = x^{n+1} \):
   \[
   y^2 - 2y + 1 = 0
   \]

10. Solving this quadratic equation:
    \[
    (y - 1)^2 = 0 \implies y = 1
    \]

11. Substituting back \( y = x^{n+1} \):
    \[
    x^{n+1} = 1
    \]

12. Since \( x \) is a real number, the solutions to \( x^{n+1} = 1 \) are:
    \[
    x = 1 \quad \text{or} \quad x = -1
    \]

13. We need to check if \( x = -1 \) satisfies the original equation. For \( x = -1 \):
    \[
    \sum_{k=1}^n \frac{k(-1)^k}{1 + (-1)^{2k}} = \sum_{k=1}^n \frac{k(-1)^k}{2}
    \]

14. When \( k \) is odd, \( (-1)^k = -1 \), and when \( k \) is even, \( (-1)^k = 1 \). Therefore, the sum does not equal \( \frac{n(n+1)}{4} \) because the terms will not sum up to the required value due to alternating signs.

15. Hence, \( x = -1 \) does not satisfy the equation. Therefore, the only solution is:
    \[
    x = 1
    \]

The final answer is \( \boxed{x = 1} \).