jmc

The first equality implies that $ab+c-bc-a=b(a-c)-(a-c)=0\Rightarrow (b-1)(a-c)=0$. By symmetry, we have: $$\begin{array}{rl}(b-1)(a-c)& =0\\ (c-1)(b-a)& =0\\ (a-1)(c-b)& =0\end{array}$$ By inspection, at least one of the following is true: $a=b$, $b=c$, or $c=a$. Without loss of generality, assume $a=b$. Substituting this into the first of our original equations, we obtain $a^2+c=ac+a\Rightarrow a^2+c=a(c+1)=41$. Since $41$ is prime and $a$ and $c$ are positive integers, either $a=1$ or $a=41$. Note that if $a=41$, then $c+1=1\Rightarrow c=0$, a contradiction with the fact that $c$ is positive. Thus, $a=b=1\Rightarrow c+1=41\Rightarrow c=40$. Therefore $a+b+c=\boxed{42}$