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 ?