$2 \cdot 32$ Let $A$ be a set of integers, where the smallest element is 1 and the largest element is 100. Except for 1, each element is equal to the sum of two elements in $A$ (which can be twice a single element). Find the minimum number of elements in set $A$.
[Solution] The set $\{1,2,4,8,16,32,36,64,100\}$ satisfies the requirement and has 9 elements, so the minimum value does not exceed 9.

If the set $A$ satisfies the requirement in the problem and has only 8 elements: $1=a_{1}<a_{2}<\cdots<a_{7}<a_{8}=100$, then because $a_{i+1} \leqslant 2 a_{i}$, we have $a_{i} \leqslant 2^{i-1}, i=1,2, \cdots, 8$. Since $a_{6} \leqslant 32, a_{7} \leqslant 64$, it follows that $a_{8}=2 a_{7}, a_{7}=50$. Also, because $a_{5} \leqslant 16, a_{6} \leqslant 32$, we have $a_{6}=25$. At this point, $a_{6} \neq 2 a_{5}$, but since $a_{4} \leqslant 8, a_{5} \leqslant 16, a_{4}+a_{5}<a_{6}$, there is a contradiction.
In summary, the set $A$ must have at least 9 elements.