Show that a system always admits a solution Consider the following system with the vector of unknowns $x\equiv (x_{y,v}: y\in \{0,1\}, v\in \mathcal{V})$, where $\mathcal{V}$ is finite and $|\mathcal{V}|\geq 2$.
$$
\begin{aligned}
& (1) \text{ }\sum_{v\in \mathcal{V}} x_{y,v}=q_y \text{ }\forall y\in \{0,1\} \\
& (2) \text{ }\sum_{v\in \mathcal{V}} x_{1,v}(z_{1,v}-z_{0,v})\geq 0 \\
& (3) \text{ }\sum_{v\in \mathcal{V}} x_{0,v}(z_{1,v}-z_{0,v})\leq 0\\
& (4) \text{ }\sum_{y\in \{0,1\}, v\in \mathcal{V}} x_{y,v}=1 \\
& (5) \text{ }0\leq x_{y,v}\leq 1 \text{ }\forall y\in \{0,1\}, \forall v \in \mathcal{V}
\end{aligned}
$$
We also know that $q_0+q_1=1$, $q_0\geq 0$, $q_1\geq 0$. I would like to show that the system admits a solution for any $q_0, q_1, (z_{1,v}, z_{0,v}: v\in \mathcal{V})$. Can you help?

How far I went: Observe that the upper bound   in (5) is implied by (4) and the lower bound in (5). Also, (4) is implied by (2). Hence, the problem reduces to:
$$
\begin{aligned}
&x_0 1_{|\mathcal{V}|}=q_0 \\
&x_1 1_{|\mathcal{V}|}=q_1 \\
&x_1 z^\top \geq 0 \\
& x_0 z^\top  \leq 0 \\
&x_0\geq 0 \\
& x_1\geq 0 
\end{aligned}
$$
where
$$ 
\begin{aligned}
&x_{y}\equiv (x_{y,v}: v \in \mathcal{V})\\
&z_{y}\equiv (z_{y,v}: v \in \mathcal{V})\\
&z\equiv z_1-z_0\\
&1_{|\mathcal{V}|}\text{ is a column vector of ones of size $|\mathcal{V}|\times 1$.}
\end{aligned}
$$
How can I proceed from here?
 A: *

*First case ($z$ can be less than $0$):

Let $\mathcal{V}=\{A\}$, and let $z_{A}=z_{1,A}-z_{0,A}=-1$
\begin{equation}
\begin{split}
x_{1,A}=& \ \ q_{1}\\
x_{0,A}=&\ \ q_{0},
\end{split}
\end{equation}
implies that $x_{1,A}\cdot z_{A}=q_{1}\cdot z_{A}<0$, so the system doesn't have solutions.


*Second case ($z_{v}\geq0,\ \forall v\in \mathcal{V}$):

Conditions (2) and (3) holds becasuse of $x_{y,v}$ being non negative. If condition (1) holds, then (4) too due to $q_{0}+q_{1}=1$. So you have to check only condition (1).
You can find $x_{y,v}$ satisfiyng condition (1) if and only if $q_{y}\leq |\mathcal{V}|$, where $|\mathcal{V}|$ denotes the number of elements in $\mathcal{V}$.


*$|\mathcal{V}|\geq 2$.

First of all,  recall that given $\lambda \in [0,1]$, $x,y\in \mathbb{R}$, then $\lambda x + (1-\lambda)y\in [x,y]$ is a convex combination of $x$ and $y$. Besides, $\sum_{i\in I}\lambda_{i}=1$ implies that $w=\sum_{i\in I}\lambda_{i}x_{i}$ is a convex combination of the points in $\{x_{i}\in \mathbb{R}:i\in I\}$, so $w\in[\min_{i\in I}(x_{i}),\max_{i\in I}(x_{i})]$.
In our problem, $z_{v}=z_{1,v}-z_{0,v}$ will play the role of $x_{i}$ and a transformation of $x_{y,v}$ the role of $\lambda_{i}$.
$$\sum_{v\in \mathcal{V}}x_{y,v}=q_{y}\iff  \sum_{v\in \mathcal{V}}\bar{x}_{y,v}=1,\ \forall y\in \{0,1\},$$ where $\bar{x}_{y,v}=x_{y,v}/q_{y}$. If $q_{y}=0$ it suffices to consider $x_{y,v}=0\ \forall v\in \mathcal{V}$.
Also we have that
\begin{equation}
\begin{split}
\sum_{v\in \mathcal{V}}x_{1,v}z_{v}\geq0 \iff&  \sum_{v\in \mathcal{V}}\bar{x}_{1,v}(q_{1}z_{v})\geq0\\
\sum_{v\in \mathcal{V}}x_{0,v}z_{v}\leq0 \iff&  \sum_{v\in \mathcal{V}}\bar{x}_{0,v}(q_{0}z_{v})\leq0
\end{split}
\end{equation}
$$\sum_{v\in \mathcal{V}}\bar{x}_{1,v}(q_{1}z_{v})\in [\min_{v\in\mathcal{V}}q_{1}z_{v},\max_{v\in\mathcal{V}}q_{1}z_{v}] ,$$
so if $\max_{v\in\mathcal{V}}(q_{1}z_{v})<0$ the system doesn't admit a solution. Reversely, suppose that $\max_{v\in\mathcal{V}}(q_{1}z_{v})\geq0$ and denote $v_{max}$ a element which maximize $z_{v}$ (it could be more), then $\bar{x}_{1,v_{max}}=1 \iff x_{1,v_{max}}=q_{1}$ and $\bar{x}_{1,v}=0\iff x_{1,v}=0, \forall v\in\mathcal{V}\setminus\{v_{max}\}$  satisfies (2).
You can proceed analogously with $y=0$ and obtain that it exists a set of $x_{0,v}$ that satisfies (3) if and only if $\min_{v\in\mathcal{V}}(q_{0}z_{v})\leq0$.
So in conclusion, the problems admits a solution if $\exists\ v,w\in \mathcal{V}$ such that $q_{0}z_{v}\leq 0\leq q_{1}z_{w}$.
