After removing any part the rest can be split evenly. Consequences? 
Let $S$ be a finite collection of real numbers (not necessarily distinct). If any element of $S$ is removed then the remaining real numbers can be divided into two collections with same size and same sum ; then is it true that all elements of $S$ are equal ?

(I know the result is true if the "reals" are replaced by "integers".)

I am not sure whether whether it is true or not. Hope this help. I think the statement is true.
I have to use your result.
Easily see that the property is true if reals is replaced by rationals.
Lemma 0: (if this is wrong, you dont need to read anymore)
If S satisfies the property, then $a S + b$ for real $a$, $b$ also satisfies the property. Here $a S$ means ${a s_1, a s_2, ...}$.
I now introduce something call relative irrational. $r_1$ and $r_2$ are relative irrational if $a r_1+b \neq r_2$ for any rational $a$, $b$. I dont know whether this is well-defined. If not, I guess my proof will fail immediately.
Claim: for any real x, there is a finite sequence of irrational $r_1$, $r_2$, $r_3$, ..., $r_n$ and a rational $q_0$ such that $x=q_1 r_1 + ... + q_n r_n+q_0$, where $q_1, ..., q_n$ are rational. For finite number of real in S, there is of course finite number of $r$. Let the number of relatively irrational used be $n$.
Then for any $s \in S$, $s=q_1 r_1 + ... + q_n r_n+q_0$ for some rationals $q_1, ..., q_n$.
When we removing arbitrary element from S, it can be divided into two collections with same size and sum. Considering their sum, we can find that the sum of coefficients of $r_k$ of one side is equal to that of the other. Also, the sum of rational of one side is equal to the other. 
We focus on the sum of rational first. It is easily to see that (or I am wrong) the rational terms of each $s \in S$ are equal, using the result given, since the collection composed by rational part of each $s \in S$ satisfies the property. Similarly, for any $r_k$, the collection composed by $q_k$ satisfies the property. Then all $q_k$ are equal in each $s \in S$.
