Odd partition with extra properties Can such a set $A=$ {$a_1,.. a_k$}  exist, such that:

*

*$\sum_i a_i = 1$ and $a_i $ are rational positive numbers

*$k$ is and odd number, and is at least $3$.

*We can partition $A$ in two parts of value $ \frac{1}{2}$ each

*$\forall a_j \in A$, 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 $\forall a_j, \quad a_j \le \frac{1}{3}$
 A: Multiply by the least common multiple $M$ of the denominators to get the equivalent problem:

*

*The $a_i$ are positive integers.

*$k$ is an odd number.

*We can partition $A$ into two parts of equal sum.

*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 $A$.
A: In fact it follows from the proof by @PeterTaylor that there are no solutions over the reals. Indeed, if we drop condition 3 then we can have solutions, $k$ is odd and all the $a_i$ are equal. But that is all.
I suspected that the following was true (But I was wrong as shown in the comments):

Suppose $a_1,a_2,\cdots,a_k$ are elements of an abelian group and , if any one is  omitted, the other $k-1$ can be split into two sets of equal sum. Then all $k$ elements are equal.

The condition 2,  that the numbers be positive is not used, only that they are not (all) zero. This means that the condition that $k$ is odd is not needed since we can just add an $a_{k+1}=0$. Also, it  is not used in the proof. The suggested example  $\{\frac13,\frac13,\frac16,\frac16\}$ does not satisfy the other conditions: one can't partition $\{\frac13,\frac13,\frac16\}$ into two parts with sum $\frac5{12}.$
We can replace condition 3 by the condition that they are not all equal.
So merely assume
1: $A=\{a_1,\cdots,a_k\}$ is a set of  reals, not all equal.
4: If we remove any element of $A$, we can partition the remaining elements $B_i=A-\{a_i\}$ into two parts of equal sum.
Then condition 4 means that there is a homogeneous system of $k$ simultaneous linear equations in $k$ unknowns, with integer coefficients (in fact just $0,\pm1$). If there are any non-trivial solutions then there are rational (and hence integer) solutions. And there could be integer solutions, the $a_i$ could all be equal with $k$ odd. But that is all: if we do add one more equation, as implied by condition $3,$ then we get a system of $k+1$ equations in $k$ variables which, as the proof shows, has no integer solutions. Hence there are no real solutions.
I am not fully satisfied with that. It seems that it should be true in any abelian group. The tags graph-theory and equitable-partition also lead one to hope for a graph-theoretic proof.
