= Solution
Choose a normalization in which the <circle group> has period $2\pi$ and the <U(1) charges> $\widehat Q_i$ of the <complex scalar fields> are integers. By <restriction of a representation>,
$$
\boxed{\widehat q_i=\widehat Q_i\pmod n.}
$$
Write $v_i=\langle\phi_i\rangle$. A <vacuum expectation value> $v_i\ne0$ is preserved by $e^{i\theta}$ precisely when $e^{i\widehat Q_i\theta}=1$. Therefore the specified <finite cyclic group> survives <spontaneous symmetry breaking> exactly when every condensing <complex scalar field> obeys
$$
\widehat Q_i\equiv0\pmod n.
$$
Only <U(1) charges> of fields with nonzero <vacuum expectation values> enter this test; in the paper all $N$ fields condense.
Suppose at least one condensing <U(1) charge> is nonzero and put $d=\gcd(|\widehat Q_i|:v_i\ne0)$. Since each such <U(1) charge> is a multiple of $d$, every $d$th <root of unity> preserves the <vacuum expectation values>. Conversely, the <Bezout identity> expresses $d$ as an integer combination of these <U(1) charges>, so a common stabilizer obeys $e^{id\theta}=1$. The <discrete remnant of a circle symmetry> is consequently
$$
\boxed{H_{\rm unbroken}=\mathbb Z_d,\qquad n_{\max}=d.}
$$
A nontrivial retained <finite cyclic group> exists exactly when $d>1$, and its possible orders are the divisors $n>1$ of $d$. For example, condensing <U(1) charges> $6$ and $10$ leave $\mathbb Z_2$. If $d=1$, the <stabilizer> is trivial. If all condensing <U(1) charges> vanish, the entire <circle group> is unbroken and there is no largest finite $n$; this case does not break the symmetry assumed in the question.
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