The soccer splitting problem in arbitrary commutative ring There's a folklore problem: 

Let $x_1, \cdots, x_{23} \in \mathbb{Z}$ be the weights of $23$ soccer players. Now Master Yoda want's to form two soccer teams with $11$ players each. Turns out for any $1 \leq i \leq 23$, one can partition $\{1, \cdots, n \} - \{ i \}$ into two disjont sets $A, B$ with $|A| = |B| = 11$ such that $\displaystyle \sum_{k \in A} x_k = \sum_{k \in B} x_k$. Prove that all numbers must be equal. 

The solution is well known and is not very hard for $\mathbb{Z}$. 
I'm wondering replacing $\mathbb{Z}$ by which commutative ring with unit $R$ makes the problem false. 
If $R = \mathbb{Q}$, then it's also same as $\mathbb{Z}$ (and the answer is affirmative), just multiply everything by the LCM of the numerators to reduce it to the case $R = \mathbb{Z}$.
If $R = \mathbb{R}$, then also the problem is true, but you need a lemma by Dirichlet (which is proven by PHP) to reduce it to the case $R = \mathbb{Z}$. 
If $R = \mathbb{C}$, then also the problem is true. Because if $\displaystyle \sum_{k \in A} z_k = \sum_{k \in B} z_k \Rightarrow \sum_{k \in A} \text{Re}(z_k) = \sum_{k \in B} \text{Re}(z_k)$, and by the previous one $R = \mathbb{R}$ applied to the real components, you get $\text{Re}(z_i) = \text{Re}(z_j)$ for all $i, j$. Similarly you prove the imaginary components are same, so all numbers are same. 
If $R = \mathbb{Q}[x], \mathbb{C}[x], \mathbb{R}[x], M_{m,n}(\mathbb{Q}),  M_{m,n}(\mathbb{Z}),  M_{m,n}(\mathbb{C}),  M_{m,n}(\mathbb{R}) $, even then the problem is true since you can look at the problem "component wise" and reduce it to the above cases.
However I have no idea whether the problem is true when $R = \mathbb{Z}_p$ for some prime $p$ or in some other rings (ring of rational functions etc). 
Is it true for all rings, or are there some rings for which this problem doesn't hold ? 
If it's false for some rings, are there any characterizations for such rings ? 
 A: As already indicated, this is a result depending only on the underlying group (rather than ring) and you can find in my earlier question, MO 105400, a comment from me that points to:

Martin, G. A. (1988). A class of Abelian groups arising from an analysis of a proof. The American Mathematical Monthly, 95(5), 433-436. JSTOR; Sci-Hub.

Here are the relevant portions:



A: Observation. If the result is true for some abelian groups $G_1$ and $G_2$ then it's also true for any extension $0 \to G_1 \to G \to G_2 \to 0$. Moreover if the result holds for $G$ then the same holds for any subgroup $H$ of $G$.
Proposition. Let $G$ be an abelian group, and let $P$ be the set of primes $p$ such that $G$ contains an element of order $p$. Then the result is true for $G$ if and only if it is true for $\mathbf{Z}/p\mathbf{Z}$ for all $p \in P$.
I'm grateful to darij grinberg for simplifying my initial proof of this proposition.
Proof. The direct implication follows from the fact that $G$ contains $\mathbf{Z}/p\mathbf{Z}$ for every $p \in P$.
Conversely, let $x_1,\ldots,x_{23} \in G$ satisfying the assumption of the problem. Let $H$ be the subgroup of $G$ generated by the $x_i$. The structure theorem of finitely generated abelian groups tells us that $H$ is a direct sum of copies of $\mathbf{Z}$ and $\mathbf{Z}/p^k\mathbf{Z}$ with $p \in P$. Since $\mathbf{Z}/p^k\mathbf{Z}$ is an iterated extension of $\mathbf{Z}/p\mathbf{Z}$, the initial observation shows that the result is true for all $\mathbf{Z}/p^k\mathbf{Z}$ with $p$ in $P$, and thus for $H$, so that all $x_i$ are equal. QED
So it remains to study the case of $\mathbf{Z}/p\mathbf{Z}$ where $p$ is prime.
Following darij grinberg's comments at this link: let $M$ be a matrix of the form $M = (\pm \delta_{i \neq j})_{1 \leq i,j \leq 23}$ where $\delta$ is the Kronecker symbol and the number of $+$ signs is 11 in each row. The upper-left $22 \times 22$ minor is odd (in particular nonzero), therefore the rank of $M$ is $22$ and its kernel is generated by the vector $(1,\ldots,1)$. This solves the original problem over $\mathbf{Z}$, and over $\mathbf{Z}/2\mathbf{Z}$.
Now over $\mathbf{Z}/p\mathbf{Z}$, the result is false if and only there exists a matrix $M$ of this form with rank $<22$, which amounts to say that the before last elementary divisor of $M$ is divisible by $p$. There are only finitely many such matrices, so the result is true if $p$ is large enough. In fact, by Hadamard's inequality, the abolute value of the $22 \times 22$ minor is bounded by $21^{11}$, so the result is true for every $p>21^{11}$.
Using Magma (I can share the code if you're interested) I generated $\approx 5 \cdot 10^7$ random matrices of this form and found that the result is false for at least 1471 values of $p$, in particular all $2<p<3529$. The largest value I got is $p=36285031$.
For a given value of $p$, it seems challenging to decide whether the result holds over $\mathbf{Z}/p\mathbf{Z}$, since one apparently needs to check all ${{22}\choose{11}}^{23}$ matrices of this form. It would be interesting to devise a better method.
