Solution:

Let $\{n\} = \{M\}(\{n\}) + \{m\}(\{n\})$, where $\{M\}(\{n\}) \geq \{m\}(\{n\})$. Put $\{M\}^{1}(\{n\}) = \{M\}(\{n\})$, $\{M\}^{2}(\{n\}) = \{M\}(\{M\}(\{n\}))$ etc. Then $\{M\}(100) \geq 50$, $\{M\}^{2}(100) \geq 25$, $\{M\}^{3}(100) \geq 13$, $\{M\}^{4}(100) \geq 7$, $\{M\}^{5}(100) \geq 4$, $\{M\}^{6}(100) \geq 2$ (and obviously $\{n\} > \{M\}(\{n\})$), so we need at least $8$ numbers. There are several ways of using $9$ numbers. For example, $\{1, 2, 4, 8, 16, 32, 36, 64, 100\}$, where $36 = 4 + 32$, $100 = 36 + 64$ and the others are double another number.

Doubling every time does not work: $1, 2, 4, 8, 16, 32, 64, 128$. But if we do not double every time, then we cannot get a number larger than $96$ with $8$ numbers: the best we can do is $1 \cdot 2^{6} \cdot (3 / 2) = 96$ (on the occasion when we do not double the best we can do is to the largest plus the next largest, or $3 / 2$ times the largest). Hence we need at least $9$ numbers. [To be more formal, write the elements as $1 = a_{1} < a_{2} < \ldots < a_{n}$, then each $a_{i}$ must be a sum of preceding elements. The largest possible $a_{i}$ is $2a_{i-1}$ and the next largest $a_{i-1} + a_{i-2}$ and so on.]