This is a follow-up question from my previous question.
Suppose there are (2n+1) vectors $\{m_1,m_2,...,m_n\}$, $\{p_1,p_2,...,p_n\}$ and $p^*$ in $R^{k+1}$. $m_i$ are weakly positive vectors. $p_i$ and $p^*$ are probability vectors (whose coordinates are weakly positive and sum to 1).
The vectors satisfy the following properties: $$ m_i p_j \geq m_i p^* = 1 \quad \forall i\neq j,$$ $$ m_i p_i = 0 \quad \forall i.$$
The question is: what is the maximum number of $n$?
This is not an answer to the question, but here are some upper/lower bounds. Firstly, if we let $A_i\subseteq\{0,1,\dots,k+1\}$ be the non zero coordinates of $m_i$, then we can't have $A_i\subseteq A_j$ for $i\neq j$, because then we would have $m_ip_j=0$ (you already mentioned this in the other question I think). This means that $(A_i)_i$ are all different and form an antichain of $\{1,\dots,k+1\}$, thus $n\leq\binom{k+1}{\left\lfloor\frac{k+1}{2}\right\rfloor}$ by Sperner's theorem.
For a superpolynomial lower bound, suppose that $\sqrt{k+1}$ is an even integer for simplicity. Let $p^*=(\frac{1}{k+1},\dots,\frac{1}{k+1})$, and for each $A\subseteq\{1,\dots,\sqrt{k+1}\}$ of size $\frac{\sqrt{k+1}}{2}$, let $p_A$ have coordinates $(p_i)_j=\frac{2}{\sqrt{k+1}}$ when $j\in A$ and $0$ if not, and let $m_A$ have coordinates $2\sqrt{k+1}$ in $\{1,\dots,\sqrt{k+1}\}\setminus A$ and $0$ else.
The number of vectors $p_A$, $m_A$ is then $\binom{\sqrt{k+1}}{\frac{\sqrt{k+1}}{2}}$, which grows faster than polynomially.
