Is there such an $n$, and if there is, how large is it, for which the sum of the first $n$ natural numbers is a three-digit number with equal digits?
1. Solution: The problem can also be stated as follows: Can the sum of the first $n$ natural numbers, $\frac{n(n+1)}{2}$, be in the form of $111k$? Therefore, we need to solve the indeterminate equation

$$
\frac{n(n+1)}{2}=111k
$$

where $k=1,2, \ldots, 9$. (1) can be rewritten as

$$
n^{2}+n-222k=0
$$

from which the positive root is

$$
n=\frac{-1+\sqrt{1+888k}}{2}
$$

Therefore, $\sqrt{1+888k}$ must be an odd integer, meaning $1+888k$ must be an odd perfect square. Since a perfect square cannot end in 7, 3, or 2, only $k=1,3,5,6,8$ are possible.

Testing these, for $k=6$,

$$
1+888k=5329=73^{2}
$$

Thus, $n=\frac{-1+73}{2}=36$. The three-digit sum is $111k=666$.

Roboz Ágnes (Bp., VI., Varga K. lg. III. o. t.)

Second Solution: (1) can also be written as

$$
n(n+1)=2 \cdot 3 \cdot 37k
$$

According to the problem,

$$
0<\frac{n(n+1)}{2}<1000
$$

or

$$
0<n^{2}<2000
$$

which means

$$
0<n<45
$$

Since in (2) the left side: $n(n+1)$ must be divisible by 37, therefore, considering (3), either $n=37$ or $n+1=37$.

If $n=37$, then $38=6k$, which means $k$ cannot be an integer.

If $n+1=37$, then $n=36=6k$ and $k=6$.

Orlik Péter (Bp., V., Eötvös g. II. o. t.)