= Solution
Use the <first-order theory> $T_{\mathbb Z}$ of a <bijection> $f$ with named inverse $g$ and no finite cycles. Its axioms are
$$
\forall x\ f(g(x))=x,\qquad\forall x\ g(f(x))=x,\qquad
\forall x\ f^n(x)\ne x\quad(n=1,2,3,\ldots).
$$
Every <first-order model> is a disjoint <union> of free two-sided <integer> orbits. Write $f^k$ for positive powers of $f$ and for powers of $g$ when $k$ is negative.
This <first-order theory> is complete and eliminates <quantifiers>. Every <first-order term> reduces to $f^k(x_i)$, and atomic <logical formulas> reduce to statements $x_j=f^k(x_i)$. To eliminate an existential <logical variable> in a <conjunction> of literals, first remove self-equalities using the no-finite-cycle axioms. If an equation fixes that <logical variable> as an <integer> translate of another <logical variable>, substitute that value. If no such equation remains, only finitely many points are excluded by the remaining inequalities, so some value exists in the infinite domain. Apply this to <disjunctive normal form> and then induct on <logical formulas>. The resulting parameter-free <first-order sentences> are decided, proving completeness.
For each fixed finite $n$, a <complete type> records a partition of the <logical variables> into common orbits and finitely many <integer> offsets within each block. There are countably many possibilities. <Countable> <first-order models> with one orbit and with two orbits are not <isomorphic>, so the <first-order theory> is not $\aleph_0$-categorical.
Thus this example has the source's explicitly defined countable-type property. In fact it also satisfies ordinary <omega-stability>: over a <countable> parameter <set>, an element is either a specified <integer> translate of a parameter, giving countably many possibilities, or lies in an entirely new orbit, giving one further <complete one-type>. The same finite-orbit-pattern description handles finite <tuples>. Therefore
$$
\boxed{T_{\mathbb Z}\text{ has countably many }n\text{-types for every finite }n,
\text{ but is not }\aleph_0\text{-categorical}.}
$$
The source prints the label “$0$-stable”; the construction meets its stated definition as well as the usual stronger <omega-stability> condition. The difference between countably many and finitely many <partial types> is exactly why Ryll-Nardzewski does not yield categoricity here.
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