Determine all real numbers a for which there exists positive reals $x_{1}, \ldots, x_{5}$ which satisfy the relations $ \sum_{k=1}^{5} kx_{k}=a,$ $ \sum_{k=1}^{5} k^{3}x_{k}=a^{2},$ $ \sum_{k=1}^{5} k^{5}x_{k}=a^{3}.$
To determine all real numbers \( a \) for which there exist positive reals \( x_1, x_2, x_3, x_4, x_5 \) that satisfy the given relations, we start by analyzing the given equations:

1. \(\sum_{k=1}^{5} kx_{k} = a\)
2. \(\sum_{k=1}^{5} k^{3}x_{k} = a^{2}\)
3. \(\sum_{k=1}^{5} k^{5}x_{k} = a^{3}\)

We will use these equations to derive a condition on \( a \).

1. **Substitute the given equations into a polynomial identity:**

   Consider the polynomial identity:
   \[
   a^2 \sum_{k=1}^{5} kx_k - 2a \sum_{k=1}^{5} k^3 x_k + \sum_{k=1}^{5} k^5 x_k = a^3 - 2a^3 + a^3 = 0
   \]

2. **Simplify the expression:**

   Substitute the given sums into the polynomial identity:
   \[
   a^2 \sum_{k=1}^{5} kx_k - 2a \sum_{k=1}^{5} k^3 x_k + \sum_{k=1}^{5} k^5 x_k = 0
   \]
   Using the given equations, we get:
   \[
   a^2 \cdot a - 2a \cdot a^2 + a^3 = 0
   \]
   Simplifying this, we have:
   \[
   a^3 - 2a^3 + a^3 = 0
   \]
   \[
   0 = 0
   \]

3. **Rewriting the polynomial identity:**

   We can rewrite the polynomial identity as:
   \[
   \sum_{k=1}^{5} k(a^2 - 2ak^2 + k^4)x_k = 0
   \]
   \[
   \sum_{k=1}^{5} k(a - k^2)^2 x_k = 0
   \]

4. **Analyzing the sum:**

   Since \( x_k > 0 \) for all \( k \), the only way for the sum to be zero is if each term in the sum is zero. Therefore, for each \( k \):
   \[
   k(a - k^2)^2 x_k = 0
   \]
   Since \( k \neq 0 \) and \( x_k > 0 \), we must have:
   \[
   (a - k^2)^2 = 0
   \]
   This implies:
   \[
   a = k^2
   \]

5. **Possible values of \( a \):**

   The possible values of \( k \) are \( 1, 2, 3, 4, 5 \). Therefore, the possible values of \( a \) are:
   \[
   a = 1^2, 2^2, 3^2, 4^2, 5^2
   \]
   \[
   a = 1, 4, 9, 16, 25
   \]

The final answer is \( \boxed{1, 4, 9, 16, 25} \).