TITLE: Self-complementary block designs
QUESTION [7 upvotes]: For what $n$ does there exist a self-complementary 
$(2n,4n-2,2n-1,n,n-1)$ balanced incomplete block design?
(All I know is that a self-complementary design with these parameters does exist for all $n$ of the form $2^k$ but doesn't exist for $n=3$.)
Motivating problem: One wants to divide $2n$ players into 2 equal-sized teams in $2n-1$ different ways in such a way that each player is on the same team as each other player exactly $n-1$ times.
For a follow-up question, see More about self-complementary block designs .

REPLY [4 votes]: These are the parameters for Hadamard 2-designs. Let $H$ be an $n\times n$ Hadamard matrix, where $n=4m$, normalized so that all entries in the first row are equal to 1. Each row apart from the first gives a partition of 
$\{1,\ldots,n\}$ into two sets of size $n/2$, hence we get
$2n-2$ blocks of size $n/2$. The resulting incidence structure is a 
2-design: because the columns of $H$ are pairwise orthogonal, any two columns differ in exactly $n/2$ positions and so any two points lie in exactly $(n-2)/2$ blocks. 
Also the designs constructed from Hadamard matrices as above are 3-designs, and it
can be show that any 3-design with these parameters arises in this way.