Suppose that ${n\choose k}, {n-1\choose k-1}, \ldots, {n-k+1\choose 1}$ are all even. (This happens for example if $k=2^\alpha-1$ and $n=2k$.) In this case, can we select ${n\choose k}/2$ sets of size $k$ from an $n$ element set such that any $i<k$ elements are contained in exactly ${n-i\choose k-i}/2$ of the selected sets?


This would be somekind of parity for the sets. If true, I am also interested in the natural generalization for $p\ne 2$ to define some $\!\!\mod p$ of a set.


The only nontrivial example that I know is for $k=3$ and $n=6$ which I have just learned from Douglas Zare who might have heard it from Robin Chapman, see [his comment here](https://mathoverflow.net/questions/163689/what-is-the-best-lower-bound-for-3-sunflowers#comment417968_163689).



Can we sometimes define the parity of a set?
I *wish* I had a real answer for you!


You are essentially interested in a tough conjecture of Hartmann, known as the "halving conjecture", which is promoted heavily by Reza Khosrovshahi. Actually, the conjecture is more general than what you are asking, since it concerns large sets of $t$-designs with arbitrary block size $k$, and I believe you have specialized to $t=k-1$. In other words, the more general conjecture asks if your condition holds for $i \le t$ when only the first $t+1$ of your binomial coefficients are even. 


Here is the closest thing I know to an open-access survey: <http://math.ipm.ac.ir/combin/publications/files/2009/PA0900076.pdf>


See Section 6 in particular.


The halving conjecture is known to be true for $2$-designs. I haven't read the proof, but it is apparently nontrivial. In any case, this means your question has an affirmative answer for $k=3$ and all admissible $n$. For general $t$, quite a few partial results are known, such as product, puncturing, and complementing constructions. There are also important connections with null designs and trades.


I hope some of this helps for your particular application(s).


