= Solution
\b[The algebra must be finite-dimensional, and there is no universal bound on its dimension.] Let $e=\varepsilon m$ and write its <coevaluation morphism> as a finite sum
$$
n(1)=\sum_{i=1}^r a_i\otimes b_i.
$$
For any $v\in A$, one <snake identity> gives
$$
v=\sum_{i=1}^r e(v,a_i)b_i.
$$
Thus the finitely many $b_i$ span $A$, making it a <finite-dimensional vector space>. The other <snake identity> shows that $e$ is <nondegenerate> in the other variable as well, so the maps to the <dual space> determined by this pairing are isomorphisms. In the standard algebraic terminology this is a <Frobenius algebra>.
Every positive finite <dimension of a vector space> occurs: take $A=k^d$ with coordinatewise multiplication and $\varepsilon(a_1,\ldots,a_d)=\sum_i a_i$. The coordinate idempotents $e_i$ give $e(e_i,e_j)=\delta_{ij}$ and $n(1)=\sum_i e_i\otimes e_i$. If zero unital algebras are allowed, the zero object gives dimension zero too.
The pairing need not make $A$ a <semisimple algebra>. For example, the <dual numbers> $k[t]/(t^2)$ with $\varepsilon(a+bt)=b$ have pairing matrix
$$
\begin{pmatrix}0&1\\1&0\end{pmatrix}
$$
and <coevaluation morphism> $1\mapsto1\otimes t+t\otimes1$. This pairing is <nondegenerate> over every <field>, despite the <nilpotent ideal> $(t)$. \b[Finite-dimensionality is the dimension conclusion; separability is not part of the given definition.]
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