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 al 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.

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.

In fact it follows from the proof by @PeterTaylor that there are no solutions over the reals.

I suspect the following is true and has an easy proof, but I do not have any proof:

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.

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 al 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.