Answer: 70

Solution:

In base $b$ the two numbers are $17_b = b + 7$ and $97_b = 9b + 7.$ We need $b + 7 \mid 9b + 7,$ and since $b + 7$ certainly divides $9(b + 7) = 9b + 63,$ this is equivalent to $$\begin{gathered} b + 7 \mid (9b + 63) - (9b + 7) \\ = 56. \end{gathered}$$

For $b \gt 9$ we have $b + 7 \gt 16,$ so $b + 7$ must be $28$ or $56,$ giving $b = 21$ or $b = 49.$ The sum is $21 + 49 = 70.$