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 $\forall a_j, \quad a_j \le \frac{1}{3}$