8. (3 points) The smallest natural number that can be expressed as the sum of 9, 10, and 11 consecutive non-zero natural numbers is


【Analysis】Let the required positive integer be $A$, then according to the problem, we have:
$$
\begin{array}{l}
A=(p+1)+(p+2)+(p+3)+\cdots+(p+9)=9 p+45, \\
A=(m+1)+(m+2)+(m+3)+\cdots+(m+10)=10 m+55, \\
A=(n+1)+(n+2)+(n+3)+\cdots+(n+9)=11 n+66,
\end{array}
$$

According to these three equations, we will discuss and solve the problem.
【Solution】Let the required positive integer be $A$, then according to the problem, we have:
$$
\begin{array}{l}
A=(p+1)+(p+2)+(p+3)+\cdots+(p+9)=9 p+45, \\
A=(m+1)+(m+2)+(m+3)+\cdots+(m+10)=10 m+55, \\
A=(n+1)+(n+2)+(n+3)+\cdots+(n+9)=11 n+66,
\end{array}
$$

where $p, m, n$ are all integers. From (1) and (2), we get: $9 p+45=10 m+55$, so
$$
9 p=10(m+1) \text{. }
$$

From (2) and (3), we get: $10 m+55=11 n+66$, so
$$
10 m=11(n+1) \text{. }
$$

Since 10 and 11 are coprime, from (5) we know that $m$ is a multiple of 11, and from (4) we know that $m+1$ is a multiple of 9, so $m$ is a multiple of 11 and leaves a remainder of 8 when divided by 9. Thus, the smallest value of $m$ is $44$, and the smallest value of $A$ is $10 \times 44+55=495$.

Answer: The smallest natural number is 495.
Therefore, the answer is: 495.