= Solution
Use one common <ultraviolet cutoff> for all physical degrees of freedom, with positive sign for each <bosonic> degree of freedom and negative sign for each <fermionic> degree of freedom. For the <mass-moment cancellation of vacuum energy divergences>, define the weighted <supertrace> $\operatorname{Str}f(m)=\sum_Bf(m_B)-\sum_Ff(m_F)$. The large-<momentum> expansion is
$$
\sum_B\sqrt{p^2+m_B^2}-\sum_F\sqrt{p^2+m_F^2}
=p\,\operatorname{Str}1+\frac{\operatorname{Str}m^2}{2p}-\frac{\operatorname{Str}m^4}{8p^3}+O(p^{-5}).
$$
After multiplication by the radial measure $p^2dp$, these three terms produce quartic, quadratic and logarithmic <ultraviolet divergences>. Therefore the necessary and sufficient cancellations for the displayed \i[unsubtracted] finite-species <vacuum energy> are
$$
\boxed{\operatorname{Str}1=0,\qquad\operatorname{Str}m^2=0,\qquad\operatorname{Str}m^4=0.}
$$
They respectively mean equal numbers of <bosonic> and <fermionic> physical degrees of freedom, equal sums of squared <masses>, and equal sums of fourth powers of <masses>. Once they hold, the remaining radial integrand falls as $p^{-3}$ and is integrable at infinity. Nonnegative <masses> introduce no lower-end divergence.
Pairing every <bosonic> degree of freedom with a <fermionic> one of the same <mass>, as in an unbroken <supersymmetric> spectrum, is sufficient and cancels the entire expression. It is stronger than these three moment conditions. For a concrete non-paired example, take bosonic squared <masses> $\{0,3,3\}m_*^2$ and fermionic squared <masses> $\{1,1,4\}m_*^2$. Both have three degrees of freedom, squared-<mass> sum $6m_*^2$ and fourth-power-<mass> sum $18m_*^4$, but their spectra differ. If a <vacuum energy> <counterterm> is instead permitted, a renormalized finite answer need not satisfy them; that is a different prescription from finiteness of the source's bare sum.
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