Solution:

We'll use $\overline{AB}$ to denote a 2 digit number with $A$ in the tens digit and $B$ in the unit digit.

Let Josie pick the number $\overline{AB}$ and Kevin $\overline{CD}$. Then we have

$$\overline{AB} = 2 \times \overline{CD}$$
$$10A + B = 20C + 2D$$

Now, $A \geq 2C > C$ so $C \neq A + B$. Therefore, $C = |A - B|$ and $D = A + B$.

- Case 1: $A \geq B$. So $C = |A - B| = A - B$. This yields
$$10A + B = 20(A - B) + 2(A + B)$$
Which simplifies to give
$$19B = 12A.$$
This can only happen when $A$ is a multiple of $19$ which is impossible since $A > 0$ is a digit.

- Case 2: $A < B$. $C = |A - B| = B - A$. This yields
$$10A + B = 20(B - A) + 2(A + B)$$
Which simplifies to give
$$4A = 3B.$$
Since $A$ and $B$ are digits (and $A$ is nonzero), this means either: $(A, B) = (3, 4)$ or $(A, B) = (6, 8)$.

To verify: $\overline{AB} = 34 \Rightarrow \overline{CD} = 17$ which works, while $\overline{AB} = 68 \Rightarrow \overline{CD} = 34$ does not work. Therefore the final answer is $34 + 17 = 51$.