= Solution
Normalize the <circle group> as in <1/d/solution> and count all <Weyl fermions> as left-handed; a right-handed <Weyl fermion> is replaced by its left-handed charge conjugate. The <restriction of a representation> gives
$$
\boxed{q_a=Q_a\pmod n.}
$$
The triangle coefficient of the <cubic U(1) gauge anomaly> is proportional to $\sum_aQ_a^3$. In a theory with this single <gauge group>, cancellation of its local pure <gauge anomaly> requires
$$
\boxed{\sum_{a=1}^M Q_a^3=0.}
$$
If absence of the <mixed gauge-gravitational anomaly> is also required, as when coupling to a general gravitational background, one additionally needs $\sum_aQ_a=0$. Additional <gauge group> factors would require their mixed <gauge anomalies> to cancel as well.
Choose integer representatives and write $Q_a=q_a+nk_a$. Expanding the cubic gives
$$
0=\sum_a Q_a^3
=\sum_aq_a^3+n\sum_a\bigl(3q_a^2k_a+3nq_ak_a^2+n^2k_a^3\bigr).
$$
Hence the inherited necessary congruence, and its gravitational counterpart when imposed, are
$$
\boxed{\sum_aq_a^3\equiv0\pmod n,\qquad
\sum_aq_a\equiv0\pmod n\quad\text{(if mixed gravitational cancellation is imposed)}.}
$$
These statements concern the same complete <fermion> spectrum used in the parent theory. They are consequences of an anomaly-free integer lift, rather than a full classification of intrinsic discrete <gauge anomalies> or a sufficient test for such a lift. They also do not imply an ordinary equality for the chosen residues: for $n=3$, the anomaly-free <U(1) charges> $(1,-1)$ have residues $(1,2)$, whose cubes sum to $9$, rather than zero. Integrating out heavy <fermions> can require extra contributions to the effective discrete <gauge anomaly> accounting.
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