Degree $2$ nilpotent matrices with non-zero product Let $n$ be sufficiently large positive integer. 
Let $M_i$ be $n$ matrices with entries in either $\mathbb{C}$
or $\mathbb{Z}/n \mathbb{Z}$.

Is it possible, (1) for $1 \le i \le n, M_i^2=0$ and
  (2) every product of the permutations of $M_i$ doesn't vanish: $\prod_{i} M_i \ne 0$?

If it is possible, $M_i$ better be as small as possible.
Explicit construction for single $n$ as large as possible is of
interest to me.

For $n=3$ solution exists with 3 by 3 matrices,
coming from parametrization of nilpotent matrices.
In flattened form:
 M1=[-59/20, -767/400, -59/400, 3, 39/20, 3/20, 20, 13, 1]
 M2=[-441, -8379, -93933, 12, 228, 2556, 1, 19, 213]
 M3=[-1931, -17379, -44413, 212, 1908, 4876, 1, 9, 23]


A general nilpotent matrix in dimension 2 can be written in the form
$$M(v):=\begin{pmatrix} xy & -x^2 \\ y^2 & -xy\end{pmatrix}, \quad v=\begin{pmatrix}x\\y\end{pmatrix}.$$ The best is to interpret $\mathfrak{sl}_2$ as $\mathfrak{sp}(V,\omega)$ where $(V,\omega)$ is a $2$-dimensional symplectic space. Then one has a natural isomorphism $\phi:V\to V^*$, and the matrix above, with suitable choice of the form, represents the map which corresponds to $v \otimes \phi(v)$ for some $v\in V$ (viewing endomorphisms as $V\otimes V^*$). Take matrices of this form with vectors $v_1$, $\ldots$, $v_n$ such that $\omega(v_i, v_j) \neq 0$ for $i\neq j$ (if you want to do that for all $n$, this requires the field to be infinite).
