What is the number of families of $4t+1$ subsets of order $2t$ of the set $\{1,2, \ldots ,p\}$, where $p$ is a prime number which is equal to $4t+1$ and the order of intersection of each pair of the subsets is $t$ or $t1$ and each subset has $t$intersection with $2t$ subsets and $t1$intersection with the other $2t$ subsets?

Yes, equivalently, the problem can be stated as follows: We are looking for the number of $p\times p$ 01matrices $A$ with all row sums $2t$ such that $AA^T=tJ+tIB$, where $B$ a 01matrix with also all row sums equal to $2t$. (Here, $J$ denotes the all1matrix). It is not hard to find circulant matrices that fulfill the conditions for each $p=4t+1$ (why only for primes?), and each of them gives rise to $(p1)!/2$ set families. But different matrices can yield the same families, e.g. for $t=1$, we can have For $t=2$: $t=3$: In fact, looking at the $t=2$ patterns that seem "complete", I'd conjecture that there are $\binom{2t}{t}$ essentially different such circulant matrices. 


I do not know the answer. However, here is part of an analysis for t=1; perhaps some of it can extend for larger t. I usually mean set of size 2t when I say set. Consider two sets with intersection C of size t. The complement of their union has size t+1, which specifies only one set which has intersection of size t1 with the two sets. This means any other set that intersects the two sets misses C. Thus (as t=1) one can specify a structure by listing the locations of successive C's. In this case, it means enumerating 5 cycles and dividing out by cyclic and reversal symmetry, giving 12 such structures. (I view it as 01 matrices where each column is proved to have 2 1's.) It may be that taking a matrix view will help with larger t. Gerhard "Ask Me About Binary Matrices" Paseman, 2013.02.23 

