\section*{Problem 1 - 141011}

Someone chooses a natural number \(n\), adds the natural numbers from 1 to \(n\) together, and obtains the sum \(1+2+\ldots+n\) as a three-digit number, which (like 777) consists entirely of the same digits.

Determine all possibilities for choosing a number \(n\) for which this is true!
}

Assuming \(n\) is a natural number that satisfies the conditions of the problem, then on the one hand,

\[
1+2+3+\ldots+(n-1)+n=\frac{1}{2} n(n+1)
\]

On the other hand, the sum that arises according to the problem can be written in the form \(111 x\), where \(x\) is a natural number such that \(1 \leq x \leq 9\). Therefore,

\[
n(n+1)=2 \cdot 111 x=2 \cdot 3 \cdot 37 x
\]

Since 37 is a prime number, it follows that either \(n\) or \(n+1\) is divisible by 37; Furthermore, since the sum \(1+2+\ldots+n\) is supposed to be a three-digit number, \(n(n+1)<2000\), and thus \(n^2<2000\) and therefore \(n<45\). Consequently, \(n\) can only be one of the numbers 36, 37.

Since \(\frac{36 \cdot 37}{2}=666\) and \(\frac{37 \cdot 38}{2}=703\), we obtain a three-digit number with three identical digits as the sum of the natural numbers from 1 to \(n\) exactly when \(n=36\).