Find the smallest $n \ge 1000$ such the sum $ 1+11+111+⋯+11⋯11 (n$ digits) is divisible by $101$ I get another training problem, for a Romanian 6th grader competition, for which I have no answer.
Find the smallest $n \ge 1000$ such the sum $ 1+11+111+⋯+11⋯11 (n digits)$ it is divisible by 101. I am familiar with the sum, I know how to compute it, but I can see how I can reason about the divisibility by $101$. 

I did a numeric analysis on it, and it looks that the solution is $n=1121$, while the first numbers for which the sum is divisible by 101 are $110,313, 403, 404, 514, 717, 807,808, 918$ . Also I did notice that $1111$ is divisible by $101$, and I also seen that the pattern $123456790$ keeps repeating on the sum.
Since $111100\ldots00$ is divisible by $101$, you can repeatedly remove the leading four $1$'s from $111\ldots 111$ until you get to one of $0,1,11,$ or $111$. And you can reduce $111$ still further to $10$. So you get the sum
$$1+11+10+0+1+11+10+0+\ldots$$

mod $101$. Can you take it from there?
Hint: In order to understand the sum, we should understand the terms that go into it. What is the $n$-digit number $11\cdots 11$ mod $101$? How many terms does it take for that sequence to repeat? Your observation that $1111\equiv 0\mod 101$ will be helpful here.