Fix $n>0$ and $X\subseteq\mathbb{R}^n$. A function $f:X\longrightarrow\mathbb{R}$ is **linear** if it is of the form
$$
f(\bar{x})=a_1x_1+\ldots+a_nx_n+b
$$
for some $a_i,b\in\mathbb{R}$.

Suppose we have linear functions $f_1,\ldots,f_t$ and $g_1,\ldots,g_t$ with the following property:

For any $\bar{x}\in X$ there are permutations $\sigma$ and $\tau$ of $\{1,\ldots,t\}$ such that
$$
f_{\sigma(1)}(\bar{x})\leq g_{\tau(1)}(\bar{x})< f_{\sigma(2)}(\bar{x})\leq g_{\tau(2)}(\bar{x})<\ldots< f_{\sigma(t)}(\bar{x})\leq g_{\tau(t)}(\bar{x}).
$$

Is it true that there are $i,j\in\{1,\ldots,t\}$ such that $f_i(\bar{x})\leq g_j(\bar{x})$ for all $\bar{x}\in X$?

When $n=1$ it's not too hard to show that the answer is yes, but the argument relies on the ordering of $\mathbb{R}$. Although in this case we only need the weaker assumption that for all $\bar{x}\in X$ and $i\in\{1,\ldots,t\}$ there are $j,k\in\{1,\ldots,t\}$ such that $f_i(\bar{x})\leq g_j(\bar{x})$ and $f_k(\bar{x})\leq g_i(\bar{x})$.

I am having trouble proving the general case, or imagining a counterexample.

It would even be helpful to show that there are $i,j\in\{1,\ldots,t\}$ and some bounded set $A$ such that $f_i(\bar{x})\leq g_j(\bar{x})$ for all $\bar{x}\in X\backslash A$.