Think big

Suppose $n$ ends in $k$ 9's (but not $k+1$ ), with $k\geq 0$ , so that $n=...a99...99$ and $n+1=...(a+1)00...00$ for some digit $0\leq a \leq 8$ . Now the difference of the digit sums of $n$ and $n+1$ , which is $9k-1$ , must be divisible by 10; the smallest solution is $k=9$ . A moment's thought will convince you that the smallest integer that satisfies the requirements is $\boxed{18999999999}$

Source

Clearly, a solution exists because $1111111110999999999$ satisfies the property, so don't input $666$ , however, it isn't the smallest solution so don't input it either.

Let $S(n)$ be the digit sum of $n$ . I claim $n$ ends in a $9$ . If it didn't, we know $S(n) \equiv 0 \mod 10$ , and so $S(n+1) \equiv S(n)+1 \equiv 1 \mod 10$ , so it couldn't be a solution. Thus, $n$ ends in a $9$ .

In fact, let $n$ end in $k$ $9$ s, for some $k\ge 1$ . Then $0\equiv S(n+1)=S(n)+1-9k \equiv 1-9k \mod 10 \Rightarrow 9k \equiv 1 \mod 10 \Rightarrow k \equiv 9 \mod 10$ .

So the least number of $9$ s $n$ must end in is $9$ . Lastly, the remaining digits must sum to be congruent to $9 \mod 10$ , so to keep the number as small as possible, we choose them to be $18$ . Therefore, the full number must be $18999999999$

...9999 + 1 = ...10000 or ...999999999 + 1 = ...1000000000 or etc. must happen for value of n plus 1 equals to value of (n+1), otherwise, an addition of 1 must not allow both value of n and value (n+1) to be both divisible by 10. The right most digit or least significant figure of integer n must be 9.

Consider 9 + 1 = 10, 9 is not divisible by 10 while 1+0 is also not divisible by 10;

Consider 19 + 1 = 20, 1+9 is divisible by 10 but 2+0 is not divisible by 10;

Consider 819 + 1 = 820, 8+1+9 is not divisible by 10 when 8+2+0 is divisible by 10;

Consider 2819 + 1 = 2820, 2+8+1+9 is divisible by 10 but 2+8+2+0 is not divisible by 10;

Consider 109 + 1 = 110, 2099 + 1 = 2100, 30999 + 1 = 31000, 409999 + 1 = 410000, 5099999 + 1 = 5100000, 60999999 + 1 = 61000000, 709999999 + 1 = 710000000 and 8099999999 + 1 = 8100000000, only n's are divisible by 10 but not (n+1)'s.

With 90999999999 + 1 = 91000000000, it satisfy! Unfortunately, this is not the smallest n.

Sum of digits of n ought to be 9 $\times$ 10 while sum of digits of (n+1) ought to be 10, for smallest n.

09|999999999 + 1 = 10|000000000 but right hand side of (n+1) makes sum of 1 not divisible by 10.

With 18999999999 + 1 = 19000000000 despite 27999999999 + 1 = 28000000000 and etc. for smallest possible, n = 18999999999 is the answer wanted. It is also found that non-continuous of 9 for right hand side of n cannot satisfy! {0, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 9, 8} are derived from {0, 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108} to suggest for $\times$ 10 above!

Summary: Sum of digits of n must be 90 rather than 0, and sum of digits of (n+1) is 10, for smallest n.

Computing method using ordinary program can hardly help for automatic verification onto big figures. This question is solved mainly by thinking. Once an upper limit was found, 666 was safely denied.

Answer: $\boxed{18999999999}$