8. In the plane region
$$
M=\left\{(x, y) \left\lvert\,\left\{\begin{array}{l}
0 \leqslant y \leqslant 2-x, \\
0 \leqslant x \leqslant 2
\end{array}\right\}\right.\right.
$$

take any $k$ points, it is always possible to divide these $k$ points into two groups $A$ and $B$, such that the sum of the x-coordinates of the points in group $A$ is no more than 6, and the sum of the y-coordinates of the points in group $B$ is no more than 6. Then the maximum value of the positive integer $k$ is . $\qquad$
8. 11.

Since there exist 12 points in the plane region $M$:
$$
\begin{array}{l}
P_{t}\left(1-10^{-4}, 1+10^{-t-3} t\right)(t=1,2, \cdots, 5), \\
P_{t}\left(1+6 \times 10^{-4}, 1-10^{-4} t\right)(t=6,7, \cdots, 12),
\end{array}
$$

these points cannot be divided into two groups as required by the problem, so $k \leqslant 11$.
Next, we prove that $k=11$ satisfies the problem's conditions.
Take any 11 points $P_{i}\left(x_{i}, y_{i}\right)$ $(i=1,2, \cdots, 11)$ in the plane region $M$. By the problem's conditions, we know $y_{i} \leqslant 2-x_{i}(i=1, 2, \cdots, 11)$. Without loss of generality, assume $0 \leqslant x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{11} \leqslant 2$, and consider the following two cases.
(1) If $\sum_{i=1}^{11} x_{i} \leqslant 6$, then take $A=\left\{P_{1}, P_{2}, \cdots, P_{8}\right\}$, $B=\left\{P_{9}, P_{10}, P_{11}\right\}$.
(2) If $\sum_{i=1}^{11} x_{i}>6$, then there exists a unique positive integer $k \in \{4,5, \cdots, 10\}$ such that $\sum_{i=1}^{k} x_{i} \leqslant 6$.
But $\sum_{i=1}^{k+1} x_{i}>6$, so $x_{k+1}>\frac{6}{k+1}$, and thus,
$$
\sum_{i=k+1}^{11} y_{i} \leqslant \sum_{i=k+1}^{11}\left(2-x_{i}\right)
$$