Janae and Jerome each have m marbles. Gwen has 4(m + 2) marbles. John has 8m + 4 marbles. If John has more marbles than Janae, Jerome, and Gwen combined, how many marbles might Janae and Jerome each have?
The total number of marbles that John, Janae, Jerome, and Gwen have is 8m + 4 + m + m + 4(m + 2)
This simplifies to 8m + 4 + m + m + 4m + 8
When simplified further, this equals 14m + 12.
Since John has more marbles than all three combined, then 14m + 12 > 2m + m + 4m + 8.
Simplifying this inequality gives 14m + 12 > 7m + 8.
Further simplifying gives 14m + 12 > 8m + 8.
After subtracting 8m from both sides of the inequality, we get 6m + 12 > 8.
Subtracting 12 from both sides of the inequality, we get 6m > -4.
Dividing both sides of the inequality by 6 gives m > -4/6.
The simplified inequality is m > -2/3.
So there is no minimum value for m since it could have any value greater than -2/3.
So if m = 1, then Janae and Jerome each have 1 marble, which is a possible number of marbles they might have. Answer: \boxed{1}.