Let $-N\leq t \leq N$.
Let $A$ be regular $(N-1)$-dimensional simplex with vertices $(t,0, \ldots, 0)\ldots (0, 0,\ldots, t)$ and $B$ be regular $(N-1)$-dimensional simplex with vertices $(t-N+1,1, \ldots, 1)\ldots (1,1, \ldots, t-N+1)$.
The intersection $A \cap B=P(N, t)$ is a polytope with $N {N-1 \choose m}$ vertices, where $m<t<m+1, \, 0<m<N, \, m \in N$. These $N {N-1 \choose m}$ vertices have $m$ coordinates $1$, $N-m-1$ coordinates $0$ and $1$ coordinate $t-m$.
Question: How to subdivide polytope $P(N,t)$ into $N {N-1 \choose m}$ equal parts, such that each part would correspond to one vertex of $P(N,t)$?
Thank you.
Here is a general way to divide a polytope into parts at each vertex. I use "face" generically for any dimension, so 0-faces are vertices, 1-faces are edges etc. We'll divide a big polytope $P$ into little polytopes $Q_v$ at each vertex $v$ of $P$. Assume that $P$ is sufficiently symmetric so that each face has an obvious center. (This is the case in your question.) Then $Q_v$ is the convex hull of the centers of each face incident to the vertex $v$. These points include the vertex $v$ itself, the midpoints of each edge at $v$, the centers of each 2-face including $v$, and so on, up to the center point of $P$.
It's easy to see that these cover the whole polytope $P$ and have disjoint interiors. For instance, use induction on the dimension.
If $P$ is vertex-transitive, then each $Q_v$ will be "equal," i.e. congruent. $P(N,t)$ is vertex-transitive in your question (any polytope whose vertices are the coordinate permutations of some point is vertex-transitive), so we have succeeded in dividing it into equal parts for each vertex.