99(1209). How many consecutive natural numbers, starting from 1, need to be added to get a three-digit number written with identical digits?
Solution. Let $1+2+3+\ldots+n=\overline{b b b}$. Then $S_{n}=$ $=\frac{(1+n) \cdot n}{2}$. We have $\frac{n(n+1)}{2}=100 b+10 b+b$, from which

$$
n(n+1)=2 \cdot 111 b, n(n+1)=2 \cdot 3 \cdot 37 b=6 b \cdot 37
$$

where $n<45(111 b<1000,222 b<2000, n<45)$.

Since the left side of the equation is the product of two consecutive natural numbers, the right side of the equation must also be the product of two consecutive natural numbers. One of them is prime (37), and the other is a multiple of 6, so it must be 36. We have $n(n+1)=36 \cdot 37, n=36$.

It is advisable to check:

$$
1+2+\ldots+36=\frac{(1+36) \cdot 36}{2}=666
$$

Answer: 36 numbers need to be added.