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".)

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 ${\bf a}=\alpha\>{\bf 1}$.

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.