2. Find the smallest natural number that can be written as the sum of 9, 10, and 11 consecutive natural numbers.
II/2. If a number can be written as the sum of 9 consecutive natural numbers, it is equal to nine times the middle number in this sequence of 9 numbers. Similarly, we conclude that it is equal to eleven times the middle number in a sequence of 11 consecutive natural numbers. Since the number can also be written as the sum of 10 consecutive natural numbers, it is equal to five times the sum of the two middle numbers in this sequence. Thus, we know that the number is divisible by 9, 11, and 5, and the smallest such number is $9 \cdot 11 \cdot 5=495$.

The middle number in the sum of 9 consecutive natural numbers is $\frac{495}{9}=55$, the middle number in the sum of 11 consecutive natural numbers is $\frac{495}{11}=45$, and the sum of the two middle numbers in the sum of 10 consecutive natural numbers is $\frac{495}{5}=99=49+50$. Thus, we have:

$$
\begin{aligned}
495 & =51+52+53+54+55+56+57+58+59 \\
& =45+46+47+48+49+50+51+52+53+54 \\
& =40+41+42+43+44+45+46+47+48+49+50
\end{aligned}
$$