Originally asked at MSE:
Let $\Sigma=\{+,<,0,1\}$ be the usual language of Presburger arithmetic. Given a "reasonable" logic $\mathcal{L}$, let $\mathbb{Pres}(\mathcal{L})$ be the $\mathcal{L}$-theory consisting of
Axioms 1-4 of the usual presentation of Presburger arithmetic, and
for each $\mathcal{L}[\Sigma]$-formula $\varphi(x,\overline{y})$, the $\mathcal{L}[\Sigma]$-sentence $$\forall \overline{y}[\varphi(0,\overline{y})\wedge\forall x(\varphi(x,\overline{y})\rightarrow\varphi(x+1,\overline{y}))\implies \forall x\varphi(x,\overline{y})].$$
For example, $\mathbb{Pres}(\mathsf{SOL})$ is fully categorical. $\mathbb{Pres}(\mathsf{FOL})$ (= usual Presburger arithmetic) is not fully categorical of course, but it is complete as an $\mathsf{FOL}$-theory: for each first-order sentence $\theta$ in the language $\Sigma$, either $\mathbb{Pres}(\mathsf{FOL})\models\theta$ or $\mathbb{Pres}(\mathsf{FOL})\models\neg\theta$.
I'm curious about whether there is a "natural" logic $\mathcal{L}$ such that $\mathbb{Pres}(\mathcal{L})$ is incomplete as an $\mathcal{L}$-theory. There are a couple natural candidates coming from a similarly-flavored analogue of this question for PA, and I think the simplest is "Ramsey logic," that is, first-order logic equipped with the Ramsey/Magidor-Malitz quantifiers (see e.g. this answer of Enayat):
Let $\mathcal{R}$ be first-order logic augmented by each Ramsey quantifier $Q^n$. Is $\mathbb{Pres}(\mathcal{R})$ complete as an $\mathcal{R}$-theory? That is, is it the case that, for every $\mathcal{R}[\Sigma]$-sentence $\theta$, either $\mathbb{Pres}(\mathcal{R})\models\theta$ or $\mathbb{Pres}(\mathcal{R})\models\neg\theta$?