6. Let the set
$$
\begin{array}{l}
M=\{1,2, \cdots, 19\}, \\
A=\left\{a_{1}, a_{2}, \cdots, a_{k}\right\} \subseteq M .
\end{array}
$$

Find the smallest $k$, such that for any $b \in M$, there exist $a_{i}, a_{j} \in A$, satisfying $b=a_{i}$ or $b=a_{i} \pm a_{j}\left(a_{i}, a_{j}\right.$ can be the same).
(Supplied by Li Shenghong)
6. By the hypothesis, in $A$, there are $k(k+1)$ possible combinations. Thus, $k(k+1) \geqslant 19$, i.e., $k \geqslant 4$.

When $k=4$, we have $k(k+1)=20$. Suppose $a_{1}<a_{2}<a_{3}<a_{4}$. Then $a_{4} \geqslant 10$.
(1) When $a_{4}=10$, we have $a_{3}=9$. At this time, $a_{2}=8$ or 7.
If $a_{2}=8$, then $20,10-9=1,9-8=1$, so it is impossible for all 19 numbers in $A$ to be generated. If $a_{2}=7$, then $a_{1}=6$ or $a_{1}=5$. Since $20,10-9=1,7-6=1$ or $20,9-7=2,7-5=2$, this is impossible.
(2) When $a_{4}=11$, we have $a_{3}=8$. At this time, $a_{2}=7$ and $a_{1}=6$, which is impossible.
(3) When $a_{4}=12$, we have $a_{3}=7$. At this time, $a_{2}=6, a_{1}=5$, which is impossible.
(4) When $a_{4}=13$, we have $a_{3}=6, a_{2}=5, a_{1}=4$, which is impossible.
(5) When $a_{4}=14$, we have $a_{3}=5, a_{2}=4, a_{1}=3$, which is impossible.
(6) When $a_{4}=15$, we have $a_{3}=4, a_{2}=3, a_{1}=2$, which is impossible.
(7) When $a_{4}=16$, we have $a_{3}=3, a_{2}=2, a_{1}=1$, which is impossible.
(8) When $a_{4} \geqslant 17$, it is impossible.
Therefore, $k \geqslant 5$.
If we take $A=\{1,3,5,9,16\}$, then $A$ satisfies the condition.
Thus, the smallest $k=5$.