In this question, rings are not necessarily finite nor do they need to be unital (i.e., the multiplicative identity may not exist). Although I shall almost exclusively discuss finite commutative unital rings, I shall not assume finiteness, commutativity, or unitality in the definitions. For a ring $R$, $0_R$ denotes the additive identity and we shall write $\text{Units}(R )$ for the set of all units in $R$.
Definition 1: A ring $R$ is said to be bekifft iff, for any collection $\left\{r_j\right\}_{j\in J}$ of integers $r_j >1$ such that $\sum_{j \in J}\,r_j=\big|R\setminus\left\{0_R\right\}\big|$, there exists a partition $\left\{S_j\right\}_{j\in J}$ of $R\setminus\left\{0_R\right\}$ such that $\left|S_j\right|=r_j$ and $\sum_{x\in S_j}\,x=0_R$ for every $j\in J$.
Definition 2: A ring $R$ is said to be betrunken iff, for any collection $\left\{r_j\right\}_{j\in J}$ of integers $r_j >1$ such that $\sum_{j \in J}\,r_j=\big|\text{Units}(R )\big|$, there exists a partition $\left\{S_j\right\}_{j\in J}$ of $\text{Units}(R)$ such that $\left|S_j\right|=r_j$ and $\sum_{x\in S_j}\,x=0_R$ for every $j\in J$.
Remark: In the case where $J$ is an infinite index set, $\sum_{j\in J}\,r_j$ is simply $|J|$.
Examples: The zero ring is vacuously bekifft and betrunken. The ring $\mathbb{Z}/3\mathbb{Z}$ is both bekifft and betrunken. (Note that bekifft-ness and betrunken-ness are the same concept for fields.) The ring $\mathbb{Z}/4\mathbb{Z}$ is not bekifft, while it is betrunken. The ring $\mathbb{Z}/9\mathbb{Z}$ is bekifft, but not betrunken. The ring $\mathbb{Z}/14\mathbb{Z}$ is neither bekifft nor betrunken.
Clearly, if a finite ring $R$ with at least three elements is bekifft, then $\sum_{x\in R}\,x=0_R$; if a ring $R$ with $\big|\text{Units}(R )\big|\geq 2$ is betrunken, then $\sum_{x\in \text{Units}(R )}\,x=0_R$. Below are three relatively easy propositions. While I am not giving a proof here, I guarantee that these propositions are not difficult.
Proposition 1: For a ring $R$, define $$\text{Ann}_R(2):=\left\{x\in R\,\big|\,2x=0_R\right\}\,.$$ If a finite ring $R$ with at least three elements is such that $\text{Ann}_R(2)$ has a nonzero sum, then $R$ is not bekifft.
Proposition 2: For every even positive integer $n>2$, $\mathbb{Z}/n\mathbb{Z}$ is not bekifft. If $\phi(n)\geq 6$ as well, then $\mathbb{Z}/n\mathbb{Z}$ is not betrunken. (Here, $\phi$ is Euler's totient function.)
Proposition 3: For an odd positive prime $p$, let $q:=p^k$ for some $k\in\mathbb{N}$ such that $q\equiv 1\pmod{3}$. Then, the field $\mathbb{F}_q$ is both bekifft and betrunken.
Remark: Proposition 2 implies that, of all even positive integers greater than $2$, only $n\in\{4,6,8,10,12\}$ are possible values such that $\mathbb{Z}/n\mathbb{Z}$ may be betrunken. It turns out that all of such rings indeed are betrunken.
Here come the main questions. Note that the answer to Question 2 is partially known (see Proposition 3).
Question 1: Let $n$ be an odd positive integer. When is the ring $\mathbb{Z}/n\mathbb{Z}$ bekifft? When is it betrunken?
Question 2: Let $F$ be a finite field. When is $F$ bekifft (equivalently, betrunken)?
Question 3: Is there an infinite ring which is bekifft? Is there a ring with infinitely many units which is betrunken?