Think Different!

Suppose $N$ ended with exactly $x$ '9's. (Clearly it must end with a 9, otherwise the digit sum will only increase by 1)

The '9's will become '0's, so the digit sum will decrease by $9x-1$ (due to the x digits being reduced by 9, and the (x+1)th digit from the right increasing by 1)

This value must be a multiple of 7. After some testing (the quickest method since the answer is less than 7) the smallest possible $x$ is 4. We know that 69999 (digit sum is 42) is possible as 70000 has digit sum 7. If there was any number smaller than that, the digit sum must be between 36 and 42, contradiction.

Hence 69999 is the smallest.

If digit sum of $N$ is $x$ , then the digits sum of $N+1$ is $x+1$

Since both $x$ and $x+1$ (in the same format) cannot be divisible by 7, we can conclude that the number $x+1$ is further added to get a number divisible by 7.

We get this when $x=42$ , since $x+1 = 43$ which when further added gives digit sum as 7 which is divisible by 7.

So we need the smallest number whose sum of the digits is 42. The number will be smallest provided each digit is as maximum as possible. Since $42 = (4\times9) +6$ , the number is 69999