jmc

Any sum formed by a combination of the numbers $2,4$ and $8$ must be divisible by $2$. The smallest possible value of such a sum is equal to $3\cdot 2=6$, and the largest possible value of such a sum is equal to $3\cdot 8=24$. After testing, we find that $$\begin{array}{r}6=2+2+2,8=4+2+2,10=4+4+2,\\ 12=4+4+4,14=8+4+2,16=8+4+4,\\ 18=8+8+2,20=8+8+4,24=8+8+8.\end{array}$$ However, we cannot find a combination that will add to be $22$: if two of the numbers are not $8$, then the maximum possible sum is $4+4+8=16$. Thus, two of the numbers picked must be $8$, but then the third ball must have the number $6$, which is not possible. Thus, the answer is the sum of the even numbers from $6$ to $24$ excluding $22$, which is $\boxed{128}$.