= Solution
Expand the <ring> <first-order language> by a new constant $c$ and consider
$$
\operatorname{Th}(\mathbb Z)\cup\{c\ne0\}\cup
\{\exists y\ (\bar n\cdot y=c):n=1,2,3,\ldots\}.
$$
Every finite <subset> is <satisfiable> in the <integer> <ring> itself. Only finitely many <divisibility> conditions occur; interpret $c$ as a nonzero common multiple of their positive <integer> indices, using $1$ if none occur. All <first-order sentences> from $\operatorname{Th}(\mathbb Z)$ are automatically true there.
<Compactness theorem> therefore gives a <first-order model>. Its reduct $R$ to the original <ring> <first-order language> satisfies the <complete theory> of $\mathbb Z$, so $R\equiv\mathbb Z$. Its element $c^R$ is nonzero and divisible by every positive standard <integer>. Thus
$$
\boxed{R\equiv\mathbb Z,\qquad c^R\ne0,\qquad
(\forall n\in\mathbb N_{>0})\ R\models\exists y\ (\bar n y=c^R).}
$$
Such an element cannot be an ordinary nonzero <integer>, since an <integer> of absolute value less than $n$ cannot be a nonzero multiple of $n$. The statement concerns each standard <integer> index separately; it does not assert <divisibility> by every internally nonzero element of $R$.
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