To begin with, in order to satisfy
$$
\prod_{i=1}^{N}\epsilon_{i}^{n_{i}}=1
$$
we must have that, for $S=\left\{ i\backepsilon\epsilon_{i}=-1\right\} $,
$$
\sum_{i\in S}n_{i}
$$
is even. Since
$$
\sum_{i=1}^{N}n_{i}=M
$$
we want to enumerate the partitions of M in N parts such that a sum
of a subset S of its parts is even. The subset is selected at random,
selecting a part if $\epsilon_{i}=-1$ and leaving a part out if $\epsilon_{i}=1$.
We don't care about the $n_{i}$ for which $\epsilon_{i}=1$.
Furthermore, in order to satisfy
$$
\sum_{i=1}^{N}\epsilon_{i}=0
$$
there needs to be the same number of $\epsilon_{i}$ such that $\epsilon_{i}=1$
as there are $\epsilon_{i}$ such that $\epsilon_{i}=-1$, so that
N has to be even.
Therefore, the partitions of M into N parts (where N is even) have
to be such that a subset of its parts is the partition of an even
number. If M is odd, this subset must be strict. And we select parts
from the partition of M at random, with a coin toss.
The problem is not solved, but it is framed in terms that could lead
to a solution.