Can such a set $A=$ {$a\_1,.. a\_k$} exist, such that:


1. $\sum\_i a\_i = 1$ and $a\_i $ are rational positive numbers
2. $k$ is and odd number, and is at least $3$.
3. We can partition $A$ in two parts of value $ \frac{1}{2}$ each
4. $\forall a\_j \in A$, let $B\_j := A - a\_j$. We can partition $B\_j$ into two groups of value $\frac{1-a\_j}{2}$.


**Examples**:  

{$\frac{1}{3} , \frac{1}{3}, \frac{1}{9}, \frac{1}{9}, \frac{1}{9}$} respect property 1, 2 4 but not 3.  

{$\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{8}, \frac{1}{8}$} respect 1, 2, 3 but not 4: removing any element yields an infeasible problem.   

{$\frac{1}{3}, \frac{1}{3}, \frac{1}{6} ,\frac{1}{6} $} respect properties 1, 3, 4, but not 2.


I suppose it should be impossible that such a partition exist. What are your thoughts?
By the way, I don't think it is important but $\forall a\_j, \quad a\_j \le \frac{1}{3}$



Odd partition with extra properties
Multiply by the least common multiple $M$ of the denominators to get the equivalent problem:


1. The $a\_i$ are positive integers.
2. $k$ is an odd number.
3. We can partition $A$ into two parts of equal sum.
4. If we remove any element of $A$, we can partition the remaining elements into two parts of equal sum.


In the rescaled problem, at least one of the $a\_i$ is odd, since otherwise we could have rescaled by $\tfrac M2$ to get integers, contradicting $M$ being the least common multiple of the denominators.


Property 3 tells us that the sum of the elements of $A$ is even. Therefore if we remove an odd $a\_i$ in step 4, each of the parts needs to sum to a non-integral total, despite being made of integers.


Therefore by contradiction there is no such set $A$.


