9. Given that the elements of set $A$ are all integers, the smallest is 1, and the largest is 200, and except for 1, every number in $A$ is equal to the sum of two numbers (which may be the same) in $A$. Then the minimum value of $|A|$ is $\qquad$ ( $|A|$ represents the number of elements in set $A$).
9. 10 .

It is easy to know that the set
$$
A=\{1,2,3,5,10,20,40,80,160,200\}
$$

meets the requirements, at this time, $|A|=10$.
Next, we will show that $|A|=9$ does not meet the requirements.
Assume the set
$$
A=\left\{1, x_{1}, x_{2}, \cdots, x_{7}, 200\right\},
$$

where $x_{1}<x_{2}<\cdots<x_{7}$ meets the requirements.
$$
\begin{array}{l}
\text { Then } x_{1}=1+1=2, x_{2} \leqslant 2+2=4, x_{3} \leqslant 8, \\
x_{4} \leqslant 16, x_{5} \leqslant 32, x_{6} \leqslant 64, x_{7} \leqslant 128 . \\
\text { By } x_{6}+x_{7} \leqslant 64+128=192<200 \text {, we know that } \\
200=x_{7}+x_{7} \Rightarrow x_{7}=100 ;
\end{array}
$$

Similarly, by $x_{5}+x_{6} \leqslant 32+64=96<100$, we know that
$$
x_{7}=100=x_{6}+x_{6} \Rightarrow x_{6}=50 \text {; }
$$

By $x_{4}+x_{5} \leqslant 16+32=48<50$, we know that
$$
x_{6}=50=x_{5}+x_{5} \Rightarrow x_{5}=25 \text {; }
$$

By $x_{3}+x_{4} \leqslant 8+16=24<25$, we know that
$$
x_{5}=25=x_{4}+x_{4} \Rightarrow x_{4}=\frac{25}{2} \text {, }
$$

which contradicts the fact that $x_{4}$ is an integer.
Thus, $|A|=9$ does not meet the requirements, i.e., $|A| \neq 9$.
Similarly, $|A| \leqslant 8$ does not meet the requirements.
In summary, the minimum value of $|A|$ is 10.