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".)

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.
