Are there specific examples of integral polyhedra that are neither Totally Unimodular nor Total Dual Integral? It is well known that if a constraint matrix $A$ is total dual integral or totally unimodular, then this is a sufficient condition of integrality of the polyhedron defined by the system $Ax \leq \beta$ (provided that $\beta$ has integer entries). But it is not a necessary condition.

Thus my question is


  Are there natural examples (from combinatorial optimization) of integral polyhedra such that their underlying constraint matrix is neither TUM nor TDI?
I'm only going to talk about the constraint matrix part of the problem, I don't know much about the linear system (TDI) part.

Call a matrix $A$ good if $\{x|Ax\leq b,x\geq 0\}$ is a integral polytope for all integral $b$.
TUM is precisely the set of good integral matrices. This is a result by Hoffman and Kruskal. see this for how the theorem comes about and the original paper.

If we relax the requirement that $A$ is integral, then we get a larger class of matrices. Call a good matrix $A$ trivial, if it is a scaling of a TUM matrix (i.e. $A=\frac{1}{k}B$ for some TUM matrix $B$ and integer $k$).

Sadly, I don't know of any natural non-trivial good matrices.

But there is a characterization of good matrices. A matrix $A$ is $1$-regular if for all non-singular square submatrix $R$, $R^{-1}$ is integral. We have the following by Appa and Kotnyek. 


  A rational matrix $A$ is good iff $A$ is $1$-regular.


Here is an example of a non-trivial good matrix.

$$\left(
\begin{array}{cc}
 \frac{1}{2} &amp; -\frac{1}{2} \\
 -\frac{1}{2} &amp; -\frac{1}{2} \\
\end{array}
\right)$$