Can a finite subset of integers ever be a maximal anti-ideal? See here for the definition of anti-ideal.
Essentially it's the definition of an ideal of $\Bbb{Z}$ (or a ring) with $\in$ replaced by $\notin$.  Let $P = \pm \Bbb{P}_o$ be $\pm$ all the odd prime numbers.
Then @quasi answered the question of whether $P$ is a maximal anti-ideal.  They proved that at least one of $\ P \cup \{2^m\}, \ P \cup \{-2^m\}, \ P \cup \{ \pm 2^m \}$ is a maximal anti-ideal for some $m$, if such $m$ exists, and if not then simply $P$ is maximal.  In other words, $P$ is at most one element away from being maximal.

Digression:
Injective $\Bbb{Z}$-module homs preserve anti-ideals.  Suppose that $f(x), f(y) \in f(A)$ where $A$ is an anti-ideal. Then $f(x - y) = f(x) - f(y) \in f(A)$, and $f^{-1}(f(x-y)) = x - y \in A$ and $x, y \in A$, a contradiction.  Similarly, for scalar multiplication.

Can a finite set $A \subset \Bbb{Z}$ ever be a maximal anti-ideal?

Digression:

Conjecture. For any proper ideal $I \subset \Bbb{Z}$ there exists an anti-ideal $A$ such that $\Delta A = \{ a - b : a, b \in A\} = I$.

 A: If a finite set $A$ is a maximal anti-ideal of $\Bbb Z$ then $A = \{1, -1\}$.
First, we note that if $A$ is an anti-ideal of $\Bbb Z$ with $1 \in A$ or $-1 \in A$ then $A \subseteq \{1, 1\}$: if some other $b \in A$ then $b \cdot 1 = b$ or $(-b) \cdot (-1) = b$ contradicts part 1 of the definition.
Therefore $\{1, -1\}$ is a maximal anti-ideal of $\Bbb Z$: given the above, we only need to check that it is indeed an anti-ideal, which is straightforward. 
Now let $A$ be a finite set which is an anti-ideal which does not contain $1$ or $-1$. Note that $0 \notin A$ since $A$ is an anti-ideal. Write $A = \{a_1, \cdots, a_n\}$. Let $p = |ra_1 \cdots a_n| + 1$, where $r \geq 1$ is such that $p > 2 \cdot \max_{1 \leq i \leq n} |a_i|$. Then $A \cup \{p\}$ is an anti-ideal also: we need to check that $p$ is not a multiple of any element of $A$, that no element of $A$ is a multiple of $p$, that $p - a \notin A$ for any $a \in A$, and that $a - b \neq p$ for any $a, b \in A$. The last three all follow from $p > 2|a|$ for all $a \in A$, while the first follows from the definition of $p = |ra_1 \cdots a_n| + 1$ and the fact that $|a| \geq 2$ for all $a \in A$.
Therefore $A$ is not a maximal anti-ideal, so any finite maximal anti-ideal must contain $1$ or $-1$, and therefore must equal $\{1, -1\}$.
