Combinatorics question: maximal number of subdivisions There is a company with $n$ workers. The company is making subdivisions, and distributes the workers among those subdivisions. Workers can work in more than one subdivision. In each subdivision, there must be an odd number of workers. The intersection of workers from two subdivisions must be even, and it can be 0 (so there can be disjoint subdivisions). What's the maximal number of subdivisions possible?
My original answer was $n$ subdivisions, each worker in a subdivision of his/her own. I'm not sure if that's correct anymore.
Thanks in advance! 

This is really not so much a combinatorics question as a question about linear algebra over the two-element field $\def\k{\Bbb F_2}\k$. Numbering the $n$ workers, each subdivision can be interpreted as a vector in $\k^n$, with coordinate $1$ in position $i$ if worker $i$ is in the subdivision, and $0$ otherwise. Making a $n\times k$ matrix $A$ with columns the vectors for each of the $k$ subdivisions, the requirements can be summarised in the matrix equation over$~\k$:
$$
  A^\top A = I_k,
$$
the $k\times k$ identity matrix over $\k$. Here the right hand side has rank$~k$, while the rank of the left hand side is at most by that of$~A$, which is at most$~n$. Therefore $k\leq n$, there cannot be more than $n$ subdivisions.
As you noted the case $k=n$ can indeed be realised, for instance by taking $A=I_k$.
