8. Given the equation $x+11 y+11 z=n\left(n \in \mathbf{Z}_{+}\right)$ has 16653 sets of positive integer solutions $(x, y, z)$. Then the minimum value of $n$ is $\qquad$
8. 2014.

Let $y+z=k$. Then the equation becomes
$$
x+11 k=n \text{. }
$$

Thus, $k=2,3, \cdots,\left[\frac{n}{11}\right]$ corresponds to equation (1) having $k-1$ solutions, where $[x]$ represents the greatest integer not exceeding the real number $x$.

If $n$ is not a multiple of 11, then the number of positive integer solutions to the original equation is
$$
\begin{array}{l}
1+2+\cdots+(m-1)=\frac{m(m-1)}{2}=16653 \\
\Rightarrow m=183 .
\end{array}
$$

In this case, the smallest $n=11 \times 183+1=2014$.
If $n$ is a multiple of 11, then the number of positive integer solutions to the original equation is
$$
\begin{aligned}
& 1+2+\cdots+(m-2) \\
& =\frac{(m-2)(m-1)}{2}=16653 \\
\Rightarrow m & =184 .
\end{aligned}
$$

In this case, the smallest $n=11 \times 184=2024$.
In summary, the smallest $n=2014$.