For an integer $n \ge 2$, find all real numbers $x$ for which the polynomial $f(x) = (x-1)^4 +(x-2)^4 +...+(x-n)^4$ takes its minimum value.
1. We start by considering the polynomial function \( f(x) = (x-1)^4 + (x-2)^4 + \cdots + (x-n)^4 \).

2. To find the critical points of \( f(x) \), we first compute its first derivative:
   \[
   f'(x) = \sum_{k=1}^n 4(x-k)^3
   \]

3. Next, we compute the second derivative of \( f(x) \):
   \[
   f''(x) = \sum_{k=1}^n 12(x-k)^2
   \]

4. Since \( n \ge 2 \), each term \( 12(x-k)^2 \) in the sum is non-negative and at least one term is positive. Therefore, \( f''(x) > 0 \) for all \( x \). This implies that \( f(x) \) is a strictly convex function, meaning it has a unique minimum.

5. To find the minimum, we solve for \( x \) such that \( f'(x) = 0 \):
   \[
   f'(x) = \sum_{k=1}^n 4(x-k)^3 = 0
   \]

6. We observe that the function \( g(x) = \sum_{k=1}^n (x-k)^3 \) is symmetric around \( x = \frac{n+1}{2} \). This is because for each \( k \) in the sum, there is a corresponding term \( (x-(n+1-k))^3 \) that balances it out when \( x = \frac{n+1}{2} \).

7. Therefore, we evaluate \( f' \left( \frac{n+1}{2} \right) \):
   \[
   f' \left( \frac{n+1}{2} \right) = \sum_{k=1}^n 4 \left( \frac{n+1}{2} - k \right)^3
   \]

8. By symmetry, the sum of the terms \( \left( \frac{n+1}{2} - k \right)^3 \) and \( \left( \frac{n+1}{2} - (n+1-k) \right)^3 \) will cancel out, leading to:
   \[
   f' \left( \frac{n+1}{2} \right) = 0
   \]

9. Since \( f'(x) \) is an increasing function (as \( f''(x) > 0 \)), \( f(x) \) decreases for \( x < \frac{n+1}{2} \) and increases for \( x > \frac{n+1}{2} \). Thus, \( f(x) \) attains its minimum value at \( x = \frac{n+1}{2} \).

The final answer is \( \boxed{ x = \frac{n+1}{2} } \).