While working with some data for the Iowa City Hospital, James got up to get a drink of water. When he returned, his computer displayed the “blue screen of death” (it had crashed). While rebooting his computer, James remembered that he was nearly done with his calculations since the last time he saved his data. He also kicked himself for not saving before he got up from his desk. He had computed three positive integers $a$, $b$, and $c$, and recalled that their product is $24$, but he didn’t remember the values of the three integers themselves. What he really needed was their sum. He knows that the sum is an even two-digit integer less than $25$ with fewer than $6$ divisors. Help James by computing $a+b+c$.
1. We are given that \(a \cdot b \cdot c = 24\) and \(a + b + c\) is an even two-digit integer less than 25 with fewer than 6 divisors.
2. First, we list the possible sets of positive integers \((a, b, c)\) whose product is 24:
   - \((1, 1, 24)\)
   - \((1, 2, 12)\)
   - \((1, 3, 8)\)
   - \((1, 4, 6)\)
   - \((2, 2, 6)\)
   - \((2, 3, 4)\)
3. Next, we calculate the sum \(a + b + c\) for each set:
   - \(1 + 1 + 24 = 26\) (not a two-digit number less than 25)
   - \(1 + 2 + 12 = 15\) (not even)
   - \(1 + 3 + 8 = 12\) (even, but we need to check the number of divisors)
   - \(1 + 4 + 6 = 11\) (not even)
   - \(2 + 2 + 6 = 10\) (even, and we need to check the number of divisors)
   - \(2 + 3 + 4 = 9\) (not even)
4. We now check the number of divisors for the sums that are even and less than 25:
   - For \(12\):
     - The divisors of 12 are \(1, 2, 3, 4, 6, 12\) (6 divisors, not fewer than 6)
   - For \(10\):
     - The divisors of 10 are \(1, 2, 5, 10\) (4 divisors, fewer than 6)
5. Since \(10\) is the only sum that meets all the criteria, we conclude that \(a + b + c = 10\).

The final answer is \(\boxed{10}\).