smc

Clearly, we can ignore the possibility that some $x_i$ are zero, as adding/removing such variables does not change the truth value of any condition, nor does it change the value of the sum of cubes. Thus we'll only consider $x_i\in \{-1,1,2\}$. Also, order of the $x_i$ does not matter, so we are only interested in the counts of the variables of each type. Let $a$ of the $x_i$ be equal to $-1$, $b$ equal to $1$, and $c$ equal to $2$. The conditions (ii) and (iii) simplify to: (ii) $2c+b-a=19$ (iii) $4c+a+b=99$ and we want to find the maximum and minimum of $2^{3}c+1^{3}b+(-1{)}^{3}a=8c+b-a$ over all non-negative solutions of the above two equations. Subtracting twice (ii) from (iii) we get $b=3a-61$. By entering that into one of the two equations and simplifying we get $c=40-a$. Thus all the solutions of our system of equations have the form $(a,3a-61,40-a)$. As all three variables must be non-negative integers, we have $3a-61\ge 0\Rightarrow a\ge 21$ and $40-a\ge 0\Rightarrow a\le 40$. For $(a,b,c)$ of the form $(a,3a-61,40-a)$ the expression we are maximizing/minimizing simplifies to $320-8a+3a-61-a=259-6a$. Clearly, the maximum is achieved for $a=21$ and the minimum for $a=40$. Their values are $M=133$ and $m=19$, and therefore $\frac{M}{m}=\boxed{7}$.