This problem comes from a problem of economics. Let $x\in [0,1]^n$. $\{x_1,x_2,\ldots,x_n\}$ is partitioned into ${S_1, S_2,\ldots,S_k}$ such that $\sum_{x_i\in S_j}x_i\leq 1$ for each $1\leq j\leq k$. This constraint can be restated as $Bx\leq 1$ componentwise, where matrix $B \in \mathbb \{0,1\}^{k \times n}$ and its component $b_{ij}=1$ if $x_j\in S_i$. For instance, suppose $x=(x_1,x_2,x_3)$ and the partition is $\{x_1\}$ and $\{x_2,x_3\}$, this constraint can be written as \begin{gather*} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 1 \end{bmatrix}\quad \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\quad \leq \begin{bmatrix} 1 \\ 1 \end{bmatrix} \end{gather*}
Let matrix $A \in \mathbb \{0,\pm 1\}^{m \times n}$ and $P = \{x\in[0,1]^n: Bx\leq 1 \text{componentwise}, Ax=c\}$ where $c\in \mathbb{Z}^m$ with each $c_i\leq 1$ (possibly negative). $P$ is known to be nonempty. What condition on $A$ would guarantee the existence of an integral vertex on $P$? From some examples I studied, my conjecture is that when $A$ is unimodular (not necessarily totally unimodular), $P$ has at least an integral vertex.