11. (9th grade) Find a natural number $n$, knowing that the sum $1+2+3+\ldots+n$ is a three-digit number consisting of identical digits.
11. $1+2+3+\ldots+n=\frac{n+1}{2} n$. This sum equals a three-digit number $\overline{a a a}$, where $\overline{a a a}=100 a+10 a+a=111 a$. We get the equation $\frac{n+1}{2} n=111 a$. From this, $a=\frac{n(n+1)}{2 \cdot 3 \cdot 37}$. Since $a$ is an integer, $n(n+1)$ must be divisible by 2, 3, and 37. This is only possible when $n=36$.