Suppose that the quotient of the division of 109 by $x$ is $q$. Since the remainder is 4, this is equivalent to $109=q x+4$ or $q x=105$. Put another way, $x$ must be a positive integer divisor of 105. Since $105=5 imes 21=5 imes 3 imes 7$, its positive integer divisors are $1,3,5,7,15,21,35,105$. Of these, 15,21 and 35 are two-digit positive integers so are the possible values of $x$. The sum of these values is $15+21+35=71$.