Question:
Is there any clean way to find the extreme points of the $n$-dimensional polytope
$$ u_{ji}\leq x_i-x_j\leq u_{ij}, ~~~~~~ \forall i< j\in [1:n]$$ $$x_i\geq 0, ~~~~~~~~~~~~~~~~~~~~~\forall i\in [1:n]$$
where $x_i$ is $i$th element of the points inside such polytope?
A Simpler One:
Could we at least find the number of such points?
Application:
In linear programming, if you have a polytope as the constraint of your points, it is only neede to evaluate the function on extreme points.
Here is a partial answer.
Let us consider :
$$\begin{cases}u_{21} &\leq &x_1-x_2&\leq &u_{12}\\ u_{31} &\leq &x_1-x_3&\leq &u_{13}\\ u_{32} &\leq &x_2-x_3&\leq &u_{23}\end{cases}\tag{1}$$
A first remark is that for any $t \in \mathbb{R}$, if $\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ belongs to the locus, all the "ray" defined by equations:
$$\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}+t\begin{pmatrix}1\\1\\1\end{pmatrix}$$
belongs to the locus (check it on system (1)). Therefore, if the locus is non void, we have a prismatic surface.
Remark: Multiplying the second sequence of inequalities by $-1$ and adding all the inequalities, we get the following necessary condition for non-vacuity:
$$u_{21}-u_{13}+u_{32} \leq 0 \leq u_{12}-u_{31}+u_{23}$$
Let us give precision about this prismatic surface. The limit cases in (1) generate the equation of 6 planes delimitating 2 by 2, 3 zones (each one delimitated by a pair of parallel planes, the intersection of these zones giving the whole shape.
Moreover, if we don't place limitations on coordinate signs, we have an unbounded prismatic volume, with (non regular) hexagonal sections (in some cases degenerated) and no vertex.
If we restrict the coordinates to be positive, we get the intersection of this prism with the first octant (green part of the figure), generating a certain number of vertices.
_Fig. : The case $u_{21}=-1;u_{12}=3;u_{31}=-2;u_{13}=5;u_{32}=-2;u_{23}=4$. The intersection of the prism with the first octant is featured in green.
If we restrict our attention to points $(a_1,a_2,a_3)$ of the respective form $(a_1,a_2,0), \ (a_1,0,a_3), \ (0,a_2,a_3)$, we will get the vertices we are looking for...