Combinatorics of sets

Let us start with $m=2$ . Let ,the partitioned set are { $a_i,a_j$ } . Let, the sum of the elements be $S$ . Then $a_i+a_j=\frac{S}{2}$ . And the some of the rest elements be $\frac{S}{2}$ . Now, if m=3 and $a_p+a_q+a_r$ = $\frac{S}{2}$ and $a_p,a_q,a_r≠a_i,a_j$ ,then obviously there is only 5 elements. $kn=5 , n=3$ !!. So, surely we have to think of rather mixed condition. How is that? Let's see. Suppose for $m=2$ the partitioned set with cardinality $m$ be { $a_i,a_j$ }. We have to proceed like this... For $m=3$ . $P_3=$ { $a_i,a_{i1},a_{i2}$ }.

Where, $a_{i} = a_{i1} + a_{i2}$ . For, $m=4$ we shall again replace a term of $P_3$ with two other terms of the rest set whose sum is equal to the former. Then how many elements do we have in the mother set? We can easily make mother set with $≤2n$ elements, but $≥n$ elements. So, $k=2$ .