jmc

If 4 is subtracted from 109, the result is 105. Then each of the two-digit numbers that will divide 109 with a remainder of 4 will divide 105 exactly. Thus, the problem is equivalent to finding all two-digit divisors of 105. Since the prime factors of 105 are 3, 5, and 7, the divisors are $3\times 5$, $3\times 7$, and $5\times 7$, or $15,21,\text{and }35$ with a sum of $\boxed{71}$.