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".)
 A: Call an $n$-tuple ${\bf a}=(a_1,a_2,\ldots,a_n)\in{\mathbb R}^n$ good if it has the property of the "collection" $S$ described in the question. Then obviously $n$ is odd. Let ${\bf 1}:=(1,1,\ldots,1)\in{\mathbb R}^n.$
Claim 0: When ${\bf a}$ is good then  the $n$-tuples
$$\lambda{\bf a}+\mu{\bf 1}\qquad(\lambda, \mu\in{\mathbb R})$$
are good as well.
Claim 1: If ${\bf a}\in{\mathbb Z}^n$ is good then ${\bf a}=a\>{\bf 1}$ for some $a\in{\mathbb Z}$.
Proof. After adding some ${\bf p}=p{\bf 1}$, $p\in{\mathbb Z}$, to ${\bf a}$ we may assume that all entries $a_k$ are nonnegative. Call $|{\bf a}|:=\sum_{k=1}^n a_k$ the norm of ${\bf a}$. We proceed by induction on the norm. When $|{\bf a}|=0$ the claim is true. Therefore assume that the claim is true for all nonnegative  $n$-tuples ${\bf a}'$ with $|{\bf a}'|<|{\bf a}|$. Since $|{\bf a}|-a_k$ is even for all $k$ it follows that either all $a_k$ are even, or all $a_k$ are odd (whence $\geq1$). In the first case ${\bf a'}:={1\over 2}{\bf a}$, and in the second case ${\bf a}':={\bf a}-{\bf 1}$, is a good nonnegative integer $n$-tuple with smaller norm. It follows that the claim holds for ${\bf a}$.
Claim 2: If ${\bf a}\in{\mathbb Q}^n$ is good then ${\bf a}=a\>{\bf 1}$ for some $a\in{\mathbb Q}$.
Claim 3: If ${\bf a}\in{\mathbb R}^n$ is good then ${\bf a}=\alpha\>{\bf 1}$ for some $\alpha\in{\mathbb R}$.
Proof. Let ${\bf a}\in{\mathbb R}^n$ be good. The set
$$V:=\left\{\sum_{k=1}^n r_k \ a_k\>\biggm|\>r_k\in{\mathbb Q}\right\}\subset{\mathbb R}$$
is a finitely generated vector space over ${\mathbb Q}$, whence finite-dimensional. Let $(e_i)_{1\leq i\leq r}$ be a basis of $V$, and let $(\phi_i)_{1\leq i\leq r}$ be the corresponding dual basis (the $r$ coordinate functionals).
Since ${\bf a}$ is good it follows by linearity that for each $i$ the $n$-tuple $\bigl(\phi_i(a_1),\phi_i(a_2),\ldots,\phi_i(a_n)\bigr)\in{\mathbb Q}^n$ is good.  Claim 2  then implies that there are numbers $q_i\in{\mathbb Q}$ $\>(1\leq i\leq r)$ with
$$\phi_i(a_k)=q_i\qquad(1\leq k\leq n)\ .$$
The latter can be rewritten as
$$a_k=\sum_{i=1}^r q_i\> e_i=:\alpha\in{\mathbb R}\qquad(1\leq k\leq n)\ ,$$
and this is tantamount to ${\bf a}=\alpha\>{\bf 1}$.
A: Yes, all elements of $S$ must be equal.
Clearly, $S$ has an odd number of elements; say $S=\{x_1,\dots,x_{2n+1}\}$. The condition "if any element of $S$ is removed then the remaining real numbers can be divided into two collections with same size and same sum" means that the column vector $\mathbf X=[x_1,\dots,x_{2n+1}]^T$ satisfies a matrix equation of the form $A\mathbf X=\mathbf0$ for some $2n+1\times 2n+1$ matrix $A$ with all diagonal entries $0$, all off-diagonal entries $\pm1$, and equal numbers of $+1$s and $-1$s on each row. All we have to do is show that such a matrix must have rank $2n$; its right null space must then be $1$-dimensional, so it contains only the constant vectors $[x,x,\dots,x]^T$. In fact, working in$\!\mod2\,$ arithmetic for simplicity, it's easy to see that the matrix can be reduced by elementary row operations to an upper triangular form with $2n$ ones and a zero on the diagonal. (Namely, swap the first two rows, and subtract them from all rows below them; swap the next two rows, and subtract them from all rows below them; and so on.)
To put it more generally, if $A$ is a square matrix of odd order $2n+1$, and if all diagonal entries are even integers, and all off-diagonal entries are odd integers, then the rank of $A$ is at least $2n$.
Alternatively, if the equation $A\mathbf X=\mathbf0$ had a nonconstant solution, it would also have a nonconstant integer solution; but the OP has already proved (see also JiK's comment on the question) the result for integers.
A: 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 $s \in S$.
