It is possible that after 8 draws, we have 2 of each of the four colors. Since the next one drawn must be one of the four colors, we will then have 3 of some color.
If, having drawn 8, we only have at most 3 colors of jelly beans, we must already have 3 of a matching color. For 3 pairs of 3 different colors is only 6 jelly beans, and we have two more, and any other combination of 6 jelly beans with 3 colors already includes 3 of one color.
It is also possible that after 8 draws, we have all four colors, but only one of some color. In this case, we have 7 jelly beans with 3 colors, and as we saw above, we can make at most 3 pairs from these, the 7th must match one of the 3 colors.
One can continue this analysis with just two colors: it can be seen that we need a mere 5 jelly beans to ensure 3 of a kind, and with only one color, we need but 3.
So no matter what, the 9th jelly bean will make (at least) the 3rd bean of some color.