Among $k$ consecutive numbers one has sum of digits divisible by $11$ Find the least positive integer $k$ with the property that given any $k$ consecutive positive integers, there is at least one whose sum of digits is divisible by $11$.
I can show for $k\leq 57$. Among any $29$ consecutive positive integers with the same hundreds' digit and higher digits, at least one has sum of digits divisible by $11$. This can be checked by casework. Since our consecutive positive integers may span two different hundreds' digits, we have $k\leq 29+28=57$.
On the other hand, among the numbers $1,2,\ldots,28$ none has sum of digits divisible by $11$, so $k\geq 29$.
 A: Denote  by $r(n)$ the remainder modulo $11$ of the decimal digit sum of $n$. A number $n$ is good, if $r(n)=0$. 
Claim: The difference $n'-n$ between two consecutive good numbers $n$, $n'$ is at most $39$, and this bound cannot be improved.
Proof:
The numbers $n:=999\,980$ and $n':=1\,000\,019$ are consecutive good numbers with $n'-n=39$. 
Now for the essential part: Denote by $p(n)\geq0$ the number of trailing nines in the decimal representation of $n$. Example: $p(4199)=2$. Then we have the  recursion
$$r(n+1)=r(n)+1+2p(n)\qquad({\rm mod}\ 11)\ ,\tag{1}$$
because in the transition $n\rightsquigarrow n+1$ the $p(n)$ trailing nines are replaced by zeros, and the immediately preceding digit is increased by $1$.
Assume now that $n$ is a good number. Write $$n=n_0+j$$ with $n_0$ divisible by $10$ and $0\leq j\leq 9$. Put $$n_k:=n_0+10k\qquad(k\geq1)\ .$$ 
If $r(n_1)\ne1$, i.e.,  $r(n_1)\in\{0,2,3,4,5,6,7,8,9,10\}$, then there is a good number $n'$ in the decade beginning with $n_1$, and $n'-n\leq19$.
If $r(n_1)=1$, but $r(n_2)\ne1$, then there is a good number $n'$ in the decade beginning with $n_2$, and $n'-n\leq29$.
It remains to consider the case  $r(n_1)=r(n_2)=1$. The recursion $(1)$ then  implies
$$10+2p(n_2-1)=0\qquad({\rm mod}\ 11)\ ,$$
or $p(n_2-1)=6\ ({\rm mod}\ 11)$. In particular $p(n_2-1)\geq2$, and this enforces $p(n_3-1)=1$. Using $(1)$ again we can conclude that $r(n_3)=2$, whence $n':=n_3+9$ is good, with $n'-n\leq39$.
