Subsets of equal size and equal sum Given a set of $n$ distinct (real) numbers and an integer $k$, what is the maximum number of subsets of size $k$ such that they all have the same sum?
Note that we are not specifying the sum. 
For example, if we have $\{1,2,...,2n+1\}$, and $k=2$, we can do $1+(2n+1)=2+2n=...$. And the maximum is $n$ 
since we can pair up those who sum to a specific number and get at most $(2n+1-1)/2$ pairs.
What happens when $k>2$? Can we get a bound? I saw similar questions asked but they are about getting an algorithm to find such subsets, and not the maximum.
 A: I interpret the question to mean: allowing overlapping subsets, and given free choice of the $n$ distinct numbers, what is the maximum number of $k$-subsets that have the same sum?
The following construction achieves asymptotically ${n/2 \choose k/2}$ such subsets.  It is probably not the maximum, and is still quite far from ${n \choose k}$, but it is a lot of subsets for large $n,k$.
Construction: Use $\{1, 2, ..., n\}$ and pair up the numbers so each pair sums to the same value $1+n$.  There are $m = \lfloor n/2 \rfloor$ such pairs, plus an extra number (the middle one) if $n$ is odd.  A $k$-subset will be made of $p=\lfloor k/2 \rfloor$ such pairs, plus an extra (fixed) number if $k$ is odd.  All such subsets obviously have the same sum.  The number of such $k$-subsets is:


*

*${m-1 \choose p}$ if $n$ is even and $k$ is odd (because in this case we need to "break" one of the $m$ pairs to provide the extra number for $k$)

*${m \choose p}$ otherwise
BTW, for small numbers we can do better.  E.g. for $n=9, k=3$, the above construction would give ${4 \choose 1} = 4$ subsets, but a magic square can give $8$ subsets, since every row sum $=$ every column sum $=$ each of the two diagonal sums $=15$.
$$\begin{align}
8 && 1 && 6\\
3 && 5 && 7\\
4 && 9 && 2
\end{align}
$$
