Harvard-MIT November Tournament

Solution:
Answer: 38

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 $10b, \ldots, 10b+9$ are all in $S$ for $b = a-1, a, a+1$. Now, let $10a$ have sum of digits $N$. Then, the sums of digits of $10a+1, 10a+2, \ldots, 10a+9$ are $N+1, N+2, \ldots, N+9$, respectively, and it follows that $n \equiv 1 \pmod{11}$.

If the tens digit of $10a$ 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 $10a$ 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.