We claim that the answer is 38. This can be achieved by taking the smallest integer in the set to be 999981. Then, our sums of digits of the integers in the set are $$45, \ldots, 53,45, \ldots, 54,1, \ldots, 10,2, \ldots, 10$$ none of which are divisible by 11.

Suppose now that we can find a larger set $S$: then we can then take a 39-element subset of $S$ which has the same property. Note that this implies that there are consecutive integers $a-1, a, a+1$ for which $10 b, \ldots, 10 b+9$ are all in $S$ for $b=a-1, a, a+1$. Now, let $10 a$ have sum of digits $N$. Then, the sums of digits of $10 a+1,10 a+2, \ldots, 10 a+9$ are $N+1, N+2, \ldots, N+9$, respectively, and it follows that $n \equiv 1(\bmod 11)$.

If the tens digit of $10 a$ is not 9, note that $10(a+1)+9$ has sum of digits $N+10$, which is divisible by 11, a contradiction. On the other hand, if the tens digit of $10 a$ is 9, the sum of digits of $10(a-1)$ is $N-1$, which is also divisible by 11. Thus, $S$ has at most 38 elements.

Motivation: We want to focus on subsets of $S$ of the form $\{10 a, \ldots, 10 a+9\}$, since the sum of digits goes up by 1 most of the time. If the tens digit of $10 a$ is anything other than 0 or 9, we see that $S$ can at most contain the integers between $10 a-8$ and $10 a+18$, inclusive. However, we can attempt to make $10(a-1)+9$ have sum of digits congruent to $N+9$ modulo 11, as to be able to add as many integers to the beginning as possible, which can be achieved by making $10(a-1)+9$ end in the appropriate number of nines. We see that we want to take $10(a-1)+9=999999$ so that the sum of digits upon adding 1 goes down by $53 \equiv 9(\bmod 11)$, giving the example we constructed previously.