Solution:
Let $n$ be the smallest number in the sequence and $m$ the smallest with last digit $0$. $m$ and $m+10$ have different digit sums unless (possibly) the penultimate digit of $m$ is $9$, but in that case $m+10$ and $m+20$ have different digit sums. So two of $m$, $m+10$, $m+20$ are sure to have different digit sums. Hence at least one has a digit sum not congruent to $1 \pmod{11}$. Adding the appropriate final digit gives a number whose digit sum is divisible by $11$. This number lies in the range $m$ to $m+29$ and $m \leq n+9$. Hence the result. $n=999981$ shows it is best possible.