This is a follow-up to the accepted answer to the question Euclidean mapping question.
We call an integral domain $R$ Euclidean if there exists a function (called a "norm") $N: R\setminus\{0\} \rightarrow \mathbb N^+$ such that for any two elements $a,b \in R\setminus \{0\}$ there exist $q,r \in R$ such that
$$a=bq+r \qquad \text{ and } \qquad r=0 \text{ or } N(r) < N(b).$$
In the linked question/answer it is shown that
$$(I) \qquad N(R) \text{ is finite } \implies R \text{ is a field}.$$
but only under the extra assumption that
$$(*) \qquad \text{ for all }x,y \in R \setminus\{0\}, \quad x \mid y \implies N(x) \le N(y)$$
(which is often tacitly assumed, and indeed one shows that if there is an $N$ in the broad sense, then one can also find an $N'$ which satisfies $(*)$).
Note that under assumption $(*)$, it's actually immediately clear that $N(x)=N(y)$ for all units $x,y \in R^\times$, which in the field case further implies that the image of $N$ is a singleton (and conversely, any such map makes any field Euclidean).
However, if we ignore assumption (*), then e.g. if $R$ is a field, actually any function $ N: R\setminus\{0\} \rightarrow \mathbb N^+$ trivially makes $R$ Euclidean (because we can always choose $r=0$), so the image of $N$ can be whatever finite or infinite set one likes. But what about the implication (I) ? In other words:
Does there exist a domain $R$ which is not a field, but is Euclidean in the broad sense with a finite set of norms?
As Qiyu Wen comments:
Given a Euclidean function $N$ on an integral domain R (in your "broad sense"), we can define
$$\tilde N(x)= \min_{y\neq 0}N(xy).$$
Then $\tilde N$ is a Euclidean function satisfying (∗). Moreover, if $N(R)$ is finite, then so is $\tilde N(R)$. See https://kconrad.math.uconn.edu/blurbs/ringtheory/euclideanrk.pdf for more details.