**Lemma 1** For any finite set $S$ of odd positive integers there is a finite multiset $A$ or real numbers such that $\sum_{\alpha \in A} \alpha^k = 0$ iff $k\in S$.

*Proof* For $S = \{k\}$ take $A_k = [1,1,(-2)^{1/k}]$.

Let $S = \{k_1,\dots, k_m\}$. Put $A = \big[\prod_{j=1}^m \alpha_j\mid \alpha_j\in A_{k_j} \text{ for } j=1,\dots,m\big]$. Then
$$
\sum_{\alpha \in A} \alpha^k = \sum_{\alpha_1\in A_{k_1}}\dots \sum_{\alpha_m\in A_{k_m}}\prod_{j=1}^m \alpha_j^k = \prod_{j=1}^m \sum_{\alpha_j\in A_{k_j}}\alpha_j^k,
$$
which is equal to zero iff one of multiplicands $\sum_{\alpha_j\in A_{k_j}}\alpha_j^k$ is zero, i.e. iff $k=k_j$ for some $j$. $\Box$

Let first $S$ be finite, and $A$ be the multiset corresponding to $S$; $|A|= m$.

Take a positive sequence $a_n$ such that $\sum_{n=1}^\infty a^k_n = \infty$ for each $k\ge 1$ (e.g. $a_n = (\log n+1)^{-1}$) and define $x_{(n-1)m+1},\dots,x_{nm}$ to be equal to $\alpha a_n$, $\alpha\in A$, in some order. Obviously, this satisfies the requirement.

**Lemma 2** For any odd positive integer $m$ there is a sequence $\{x_n,n\ge 1\}$ or real numbers such that the sequence $\{z_1^m+\dots+z_n^m,n\ge 1\}$ is unbounded, while $\sup_{k\neq m,n\ge 1}|z_1^k+\dots+z_n^k|<\infty$.

*Proof* Use the above construction with $S=\{1,2,\dots,m-1\}$, $a_n = n^{-1/m}$, to get $\{z_n,n\ge 1\}$. Then $\{z_1^k+\dots+z_n^k,n\ge 1\}$ is obviously bounded for $k<m$, unbounded for $k=m$, and $$\sup_{k> m,n\ge 1}|z_1^k+\dots+z_n^k| \le C \sum_{n=1}^{\infty} n^{-(m+1)/m}<\infty, $$
as required. $\Box$

Now let $S$ be arbitrary set of odd positive integers, $T$ be its complement and $\{k_n,n\ge 1\}$ be a sequence of integers from $T$ such that each integer from $T$ appears in it infinitely often.

Using Lemma 2, for each $n\ge 1$, we can construct a sequence $\{y_j(n),j=1,\dots,m_n\}$ such that $\sum_{j=1}^{m_n} y_j(n)^{k_n}\ge 1$ and $\sup_{k\neq k_n,j=1,\dots,m_n}|y_1(n)^k+\dots+y_j(n)^k|<2^{-n}$. Setting \begin{align}
& x_i = y_i(1), &1\le i\le m_1,\\ & x_i = y_{i-m_1}(2), &m_1<i\le m_1+m_2,\\ & x_i = y_{i-m_1-m_2}(3), &m_1+m_2<i\le m_1+m_2+m_3
\end{align}
etc, we get the required series. Moreover, $|\sum_{n=1}^\infty x_n^k|\le 1$ for $k\in S$, and the series converges uniformly in $k\in S$.