Solution. According to Abel formula, we note that
$$\begin{aligned} \sum_{i=1}^{n} y_{i}\left(x_{i}^{13}-x_{i}\right)= & \left(y_{1}-y_{2}\right)\left(x_{1}^{13}-x_{1}\right)+\left(y_{2}-y_{3}\right)\left(x_{1}^{13}+x_{2}^{13}-x_{1}-x_{2}\right)+\ldots \\ & +\left(y_{n-1}-y_{n}\right)\left(\sum_{i=1}^{n-1} x_{i}^{13}-\sum_{i=1}^{n-1} x_{i}\right)+y_{n}\left(\sum_{i=1}^{n} x_{i}^{13}-\sum_{i=1}^{n} x_{i}\right) \end{aligned}$$

Since $y_{k} \leq y_{k+1} \forall k \in\{1,2, \ldots, n-1\}$, we only need to prove that
$$\sum_{i=1}^{k} x_{i}^{13} \geq \sum_{i=1}^{k} x_{i} \Leftrightarrow \sum_{i=1}^{k} x_{i}\left(x_{i}^{12}-1\right) \geq 0$$

Applying Abel formula again, we have
$$\begin{aligned} \sum_{i=1}^{k} x_{i}\left(x_{i}^{12}-1\right)= & \left(x_{1}-x_{2}\right)\left(x_{1}^{12}-1\right)+\left(x_{2}-x_{3}\right)\left(x_{1}^{12}+x_{2}^{12}-2\right)+\ldots \\ & +\left(x_{k-1}-x_{k}\right)\left(\sum_{i=1}^{k-1} x_{i}^{12}-k+1\right)+x_{k}\left(\sum_{i=1}^{k} x_{i}^{12}-k\right) \end{aligned}$$

Notice that $x_{i} \in[-1,1], \forall i \in\{1,2, \ldots, n\}$ so $\sum_{i=1}^{j} x_{i}^{12} \leq j \forall j \in\{1,2, \ldots, k\}$. Moreover, because $x_{1} \leq x_{2} \leq \ldots \leq x_{k}$, every term in the above sum except the last term is non-negative. If $x_{k} \leq 0$, we are done. Otherwise, suppose that $x_{k} \geq 0$, then $x_{i} \geq 0 \forall i \geq k+1$. This implies (by hypothesis)
$$\sum_{i=k+1}^{n} x_{i}^{13} \leq \sum_{i=k+1}^{n} x_{i} \Rightarrow \sum_{i=1}^{k} x_{i}^{13} \geq \sum_{i=1}^{k} x_{i}$$