Your conclusion is correct. Here is a proof:

Let N be a two-digit integer, say N = 10a + b, where a and b are digits (0 ≤ a ≤ 9, 1 ≤ b ≤ 9). Then 10N - N = 9N = 90a + 9b.

The sum of the digits of 9N is equal to the sum of the digits of 90a + 9b, which is equal to 9(a + b). Thus, we need to find all two-digit integers N such that 9(a + b) is divisible by 170.

Since 170 is not divisible by 9, it follows that a + b must be divisible by 170. But the maximum possible value of a + b is 18, and 170 is not a factor of any number between 10 and 18. Therefore, there is no two-digit integer N for which the sum of the digits of (10N - N) is divisible by 170.