Suppose $\{x_i\}$ satisfies the conditions of the problem, then
$$\sum_{k=1}^{n} kx_{k} = a, \quad \sum_{k=1}^{n} k^{3}x_{k} = a^{2}, \quad \sum_{k=1}^{n} k^{5}x_{k} = a^{3}.$$
由歌西不等式得 By the Cauchy-Schwarz inequality, we get
$$\begin{aligned} aa^3 &= \left(\sum_{k=1}^{n} kx_k\right) \left(\sum_{k=1}^{n} k^5 x_k\right) \\ &\geq \left(\sum_{k=1}^{n} k^3 x_k\right)^2 = a^4.\end{aligned}$$
Therefore equality must hold in the above inequality. Hence we should have:
(1) If none of the $x_k$ is $0$, then
$$\frac{1}{k^4} = \frac{kx_k}{k^5 x_k} \text{ is constant}$$
Since the value of $k$ is not fixed, this is clearly impossible.
(2) If among $x_k(1 \le k \le n)$ there is a $0$, without loss of generality assume $x_1 = 0$. Then
$$\sum_{k=2}^{n} kx_{k} = a, \quad \sum_{k=2}^{n} k^{3}x_{k} = a^{2}, \quad \sum_{k=2}^{n} k^{5}x_{k} = a^{3}.$$
For the above conditions, applying the Cauchy-Schwarz inequality again, we obtain: among $x_2, x_3, \dots, x_n$ there is a $0$.
Repeating the above steps, we obtain the following conclusion:

Among $\{x_k\}$ at most one is nonzero, let it be $x_i(x_i \neq 0, i$ can take the value $1, 2, \dots, n$), then
$$ix_i = a, \quad i^3 x_i = a^2, \quad i^5 x_i = a^3,$$
thus, $i^2 = a$. From this we know: all possible values of $a$ are: $1, 2^2, 3^2, \dots, n^2$.