Suppose $(x_n)_{n=1}^{k}\ $ is a decreasing $k$-tuple of positive real numbers. Let $(y_n)_{n=1}^{2k}\ $ be two copies of $(x_n)_{n=1}^{k},\ $ that is, $$ y_n= \begin{cases} x_n&\text{if}\ 1\leq n \leq k \\ x_{n-k}&\text{if}\ k+1 \leq n \leq 2k.\\ \end{cases} $$
Does there exist a permutation $(\sigma(y_n))_{n=1}^{2k}\ $ of $(y_n)_{n=1}^{2k}\ $ and an increasing tuple of $k$ integers $0\leq m_1 < m_2 < \ldots <m_k\leq 2k,\ $ such that the $k$-tuple $(z_n)_{n=1}^{k}:= \displaystyle\left(\sum_{i=m_n+1}^{m_{n+1}} \sigma(y_{i})\right)_{n=1}^{k} $ is convex decreasing and $z_n\geq x_n\ $ for all $i\in\{1,2,\ldots\ n\}?$ [Convex means $x_n-x_{n+1}\geq x_{n+1}-x_{n+2}\ \forall\ n\in\{1,2,\ldots\ n-2\}].$
The question arose when thinking about this answer to my previous question, which I admit I do not fully understand, but maybe that answer answers this question, maybe not? Even if it does, I welcome other methods to answer the question posed above, or a counter-example.
My thoughts are vague, like the greedy algorithm or methods related to Cauchy's Condensation test could be useful somehow.
According to your definition of $(z_n)_{n=1}^{k}$, we have $z_k=\sum_{i=m_k+1}^{m_{k+1}} \sigma(y_{i})$, but $m_{k+1}$ is undefined. We fix this by assuming additionally that $m_k<m_{k+1}\leq 2k$.
This answer is an elaboration of Adam's comment, so we shall follow it. We can even relax the condition $z_n\geq x_n$ for each natural $n\le k$.
For each natural $n\le k$ put $t_k=\lfloor z_k/M^{k-1}\rfloor$. Then the sequence $(t_n)_{n=1}^{k}$ is nonincreasing. For each natural $n<k$ we have $$M^{k-1}<x_n\le\sum_{i=0}^{k-1} M^i=M^{k-1}+\frac{M^{k-1}-1}{M-1}<M^{k-1}\left(1+\frac{1}{M-1}\right).$$
Now we assume that $M\ge k+2$. Then for any natural $s\le k+1$ and any sum $S$ of $s$ such $x_n$'s we have $sM^{k-1}<S<(s+1)M^{k-1}$. Then for each natural $n\le k$, $z_n=\sum_{i=m_n+1}^{m_{n+1}} \sigma(y_{i})$ and $t_n$ is the number of summands in this sum which are at least $M^{k-1}$. It implies that $\sum_{n=1}^k t_k\le 2k-2$. Therefore there exists $N=\min\{n:t_n\le 1\}$.
The sequence $(t_n)_{n=1}^{k}$ has at most two zeroes. Suppose that $N\le k-2$.Then $t_N=\dots=t_{k-2}=1$ and for each natural $n$, with $N\le n\le k-2$, $z_n$ equals $x_i$ for some $i<k$, and in the sequence $(z_n)_{n=N}^k$ there are no three equal members. Then there exists $m\in \{k-2,k-1\}$ such that $z_m\ne z_{m+1}$. Suppose for a contradiction that $N\le k-3$. Since the sequence $(z_n)_{n=1}^{k}$ is nonincreasing, there exist natural $i\le j<k$ such that $z_{m-1}=1+\sum_{s=i}^{k-1} M^s$ and $z_m=1+\sum_{s=j}^{k-1} M^s$. Since $z_m\ne z_{m+1}$, $$z_m-z_{m+1}\ge M^j-1>\frac{M^j-1}{M-1}=\sum_{s=0}^{j-1} M^s\ge \sum_{s=i}^{j-1} M^i=z_{m-1}-z_m,$$ a contradiction. Thus $N\ge k-2$.
Now we assume that $M\ge k+3$. If $t_{N-2}=t_{N-1}$ then $$(k+1)\sum_{i=0}^{k-2} M^i>z_{N-2}-z_{N-1}\ge z_{N-1}-z_N>2M^{k-1}-\sum_{i=0}^{k-1} M^i,$$ so $$\frac{(k+2)(M^{k-1}-1)}{M-1}>M^{k-1}$$ and thus $k+2>M-1$, a contradiction. Therefore $t_n\ge 3$ for each natural $n\le N-2$.
Thus $$2k-2\ge \sum_{n=1}^k t_k\ge (N-2)3\ge (k-4)3,$$ so $k\le 10$ and we have a counterexample for bigger values of $k$.