Solution:

Let Chloe be $n$ today; Zoe is $1.$ In $t$ years her age over Zoe's is $\frac{n + t}{1 + t} = 1 + \frac{n - 1}{1 + t},$ an integer exactly when $1 + t$ divides $n - 1.$ So the number of such birthdays is the number of divisors of $n - 1.$ Nine of them means $n - 1$ has $9$ divisors, forcing $n - 1 = 2^2 \cdot 3^2 = 36$ (the only two-digit choice). So Chloe is $37$ and Joey is $38.$ Now Joey's age $38 + t$ is a multiple of $1 + t$ iff $1 + t$ divides $37.$ The next such time is $t = 36,$ when Joey is $74.$ Its digit sum is $7 + 4 = 11.$ Thus, E is the correct answer.