The wikipedia article lists two. The second one is related to Godel's second incompleteness theorem.
"Let # be a standard Gödel numbering for the formulas of Peano arithmetic. Then the set A = { #(ψ) : PA ⊢ ψ} of provable formulas and the set B = { #(ψ) : PA ⊢ ¬ψ} of refutable formulas are recursively inseparable. The inseparability of the sets of provable and refutable formulas holds for many other formal theories of arithmetic (Smullyan 1958)."
Yes, there are lots of these - and the relevant notion is the arithmetic hierarchy.
There's a potential point of confusion worth addressing here, especially since it appeared in a now-deleted answer: we cannot conflate "true in $\mathbb{N}$" with "provable in $\mathsf{PA}$." In particular, for each formula $\varphi$ the set $$\{x: \mathsf{PA}\vdash\varphi(x)\}$$ is indeed r.e., but that does not mean that the set $$\{x: \mathbb{N}\models\varphi(x)\}$$ need be anywhere close to r.e.
Here's a brief summary. Every formula $\psi$ in the language of arithmetic is ($\mathsf{PA}$-provably) equivalent to one of the form $$Q_1x_1Q_2x_2....Q_nx_n\varphi(x_1,...,x_n)$$ where each $Q_i$ is a quantifier (either $\forall$ or $\exists$) and $\varphi$ uses only bounded quantifiers (quantifiers of the form $\forall y<n$ and $\exists y<n$). We can get an upper bound on the complexity of the set defined by $\psi$ by looking at:
the outermost quantifier $Q_1$, and
the number of quantifier alternations ("$\forall\exists$" or "$\exists\forall$" - this can be much less than the total number of quantifiers).
A formula of the above type whose outer quantifier is $\exists$ and which has $i$-many quantifier alternations is called $\Sigma_{i+1}$; a formula whose outer quantifier is $\forall$ and which has $i$-many quantifier alternations is called $\Pi_{i+1}$.
The first part of the connection between the arithmetic hierarchy and computational complexity is the following:
A set is r.e. iff it is definable by a $\Sigma_1$ formula.
See here. More generally, there is a connection between the arithmetical hierarchy on the one hand and Turing reducibility and the Turing jump and :
For each $n\in\mathbb{N}$ we have $X\le_T{\bf 0^{(n)}}$ iff $X$ is definable by a $\Sigma_{n+1}$ formula and $X$ is definable by a $\Pi_{n+1}$ formula.
This is due to Post. You may also find Shoenfield's limit lemma to be of interest. The simplest natural example of an arithmetic set which is not r.e., or indeed Turing-equivalent to any r.e. set (to rule out things like "the complement of the halting problem") is in my opinion the set of indices for Turing machines which halt on all inputs. This set, which is often denoted "$\mathsf{Tot}$" (for "total"), has Turing degree ${\bf 0''}$ and is $\Pi_2$ but not $\Sigma_2$.
(We say that a set is $\Sigma_n$ if it is definable by a $\Sigma_n$ formula, and similarly for $\Pi_n$; additionally, we say that a set is $\Delta_n$ iff it is both $\Sigma_n$ and $\Pi_n$. Note that there is no such thing as a "$\Delta_n$ formula" - whereas $\Sigma_n$-ness and $\Pi_n$-ness are syntactic properties, $\Delta_n$-ness is genuinely "semantic.")
Another important class of examples comes from the notion of bounded truth predicates. Per Tarski, the set $$\{\ulcorner\psi\urcorner: \mathbb{N}\models\psi\}$$ is not arithmetical (here "$\ulcorner\cdot\urcorner$" is your favorite Godel numbering function). However, for each $n$ the set of Godel numbers of true $\Sigma_n$ sentences is indeed $\Sigma_n$. The set of Godel numbers for false $\Sigma_n$ sentences (or Turing-equivalently the set of Godel numbers of true $\Pi_n$ sentences) is therefore $\Pi_n$ but not $\Sigma_n$. Similarly, the set of Godel numbers of false $\Pi_n$ sentences (or Turing-equivalently the set of Godel numbers of true $\Sigma_n$ sentences) is $\Sigma_n$ but not $\Pi_n$. Now just pick some really big $n$ (that is, $n>1$).
Best Answer
I don't think you can avoid bringing up partial functions, as the sets $A$ and $B$ cannot be recursive, and so they cannot be the domain of total functions.
Let $A$ be the set of $x$ such that $\varphi_{x}(x) = 0$ (so that the relation holds and is defined) and let $B$ be the set of $x$ such that $\varphi_{x}(x) = 1$. These sets are semirecursive, so recursively enumerable, and disjoint by construction. Suppose a recursive set $C$ existed satsifying the above-mentioned qualities. Then we can let $\chi$ be its characteristic function, and let $e$ code for $\chi$. We now ask: is $e$ in $C$? If $e$ is in $C$, then
$$\varphi_{e}(e) = 1$$
So that $e \in B$, which is impossible. Conversely, if $e$ is not in $C$, then
$$\varphi_{e}(e) = 0$$
But then $e \in A$, and hence in $C$, which gives a contradiction.