Let $ M = \{a_1,a_2,\ldots,a_n\}$ such that $ 1 = a_1 < a_2 < \ldots < a_n = 100$. Lemma 1: $ a_k \leq 2^{k-1}$. Proof: By induction. $ k = 1$ is true since $ a_1 = 1 \leq 2^0$. Suppose that the lemma is true for $ k$. Then $ a_k \leq 2^{k-1}$. Suppose $ a_{k+1}$ is the sum of $ a_i$ and $ a_j$ with $ a_i \leq a_j$. Then $ a_i \leq a_j < a_{k+1}$. Thus $ a_i \leq a_j \leq a_k$, so $ a_{k+1} = a_i + a_j \leq 2a_k \leq 2 \cdot 2^{k-1} = 2^k$, establishing case $ k+1$. Lemma 2: For $ k > 1$, $ 3 \cdot 2^{k-3} < a_k \leq 2^{k-1} \Rightarrow a_{k-1} = a_k/2$. That is, $ a_k$ is even. Proof: We have $ a_k = a_i + a_j$ for some $ i,j$. WLOG $ a_i \leq a_j$. Since both are positive, $ a_i \leq a_j < a_k$. Suppose $ a_i < a_{k-1}$. Then $ a_i \leq a_{k-2} \leq 2^{k-3}$ by Lemma 1. Also, $ a_j \leq a_{k-1} \leq 2^{k-2}$ by Lemma 1 again, so $ a_i + a_j \leq 2^{k-3} + 2^{k-2} = 3 \cdot 2^{k-3} < a_k$, contradiction. Thus $ a_i = a_j = a_{k-1}$, so $ a_k = 2a_{k-1}$. Lemma 3: $ 3 \cdot 2^{k-3} < a_k \leq 2^{k-1} \Rightarrow a_k = 2^{k-1}$. Proof: By induction. $ k = 1$ is obvious since $ a_1 = 1 = 2^0$. Suppose true for $ k$, and suppose $ 3 \cdot 2^{k-2} < a_{k+1} \leq 2^k$. By Lemma 2, $ a_{k+1} = 2a_k$. But $ a_k = 2^{k-1}$ by the inductive hypothesis. Therefore $ a_{k+1} = 2^k$, and the lemma is proven. Because $ a_n = 100$ and $ a_n \leq 2^{n-1}$ by Lemma 1, $ n \geq 8$. If $ n = 8$, then $ a_8 = 100$ and thus $ 3 \cdot 2^5 < a_8 \leq 2^7$. But $ 100 \neq 2^7$, contradicting Lemma 3. So $ n \geq 9$. The following set works for $ n = 9$: $ \{1,2,4,8,16,32,64,96,100\}$.