For the given natural number $n(n>$ 5), find all real numbers $a$ such that there exist non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying
$$
\sum_{k=1}^{n} k x_{k}=a, \sum_{k=1}^{n} k^{3} x_{k}=a^{2}, \sum_{k=1}^{n} k^{5} x_{k}=a^{3} .
$$
Assume $\left\{x_{i}\right\}$ satisfies the conditions, then,
$$
\sum_{k=1}^{n} k x_{k}=a, \sum_{k=1}^{n} k^{3} x_{k}=a^{2}, \sum_{k=1}^{n} k^{5} x_{k}=a^{3} .
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
a a^{3} & =\left(\sum_{k=1}^{n} k x_{k}\right)\left(\sum_{k=1}^{n} k^{5} x_{k}\right) \\
& \geqslant\left(\sum_{k=1}^{n} k^{3} x_{k}\right)^{2}=\left(a^{2}\right)^{2}=a^{4} .
\end{aligned}
$$

Therefore, the equality in the above inequality holds. Hence, we have:
(1) If $x_{k}$ are all non-zero, then $\frac{1}{k^{4}}=\frac{k x_{k}}{k^{5} x_{k}}$ is a constant. Since $k$ varies, this is clearly impossible.
(2) If $x_{k}(1 \leqslant k \leqslant n)$ contains 0, without loss of generality, let $x_{1}=0$. Then,
$$
\sum_{k=2}^{n} k x_{k}=a, \sum_{k=2}^{n} k^{3} x_{k}=a^{2}, \sum_{k=2}^{n} k^{5} x_{k}=a^{3} \text {. }
$$

For the above three conditions, using the Cauchy-Schwarz inequality again, we similarly conclude that $x_{2}, x_{3}, \cdots, x_{n}$ contain 0.

Repeating the above steps, recursively, we conclude: $\left\{x_{k}\right\}$ contains at most one non-zero. Let it be $x_{i}\left(x_{i} \neq 0, i\right.$ can be $\left.1,2, \cdots, n\right)$. Then,
$$
i x_{i}=a, i^{3} x_{i}=a^{2}, i^{5} x_{i}=a^{3} .
$$

Thus, $i^{2}=a$.
It follows that all possible values of $a$ are $1,2^{2}, 3^{2}, \cdots, n^{2}$.