The set $M$ consists of integers, the smallest of which is $1$ and the greatest $100$. Each member of $M$, except $1$, is the sum of two (possibly identical) numbers in $M$. Of all such sets, find one with the smallest possible number of elements.
1. **Define the set \( M \):**
   Let \( M = \{a_1, a_2, \ldots, a_n\} \) such that \( 1 = a_1 < a_2 < \ldots < a_n = 100 \).

2. **Lemma 1: \( a_k \leq 2^{k-1} \):**
   - **Base Case:** For \( k = 1 \), \( a_1 = 1 \leq 2^0 \).
   - **Inductive Step:** Assume \( a_k \leq 2^{k-1} \) for some \( k \). We need to show \( a_{k+1} \leq 2^k \).
     - Since \( a_{k+1} \) is the sum of two elements in \( M \), say \( a_i \) and \( a_j \) with \( a_i \leq a_j \), we have \( a_i \leq a_j < a_{k+1} \).
     - Thus, \( a_i \leq a_j \leq a_k \), so \( a_{k+1} = a_i + a_j \leq 2a_k \leq 2 \cdot 2^{k-1} = 2^k \).
     - This completes the inductive step, proving \( a_k \leq 2^{k-1} \).

3. **Lemma 2: For \( k > 1 \), \( 3 \cdot 2^{k-3} < a_k \leq 2^{k-1} \Rightarrow a_{k-1} = \frac{a_k}{2} \):**
   - Given \( a_k = a_i + a_j \) for some \( i, j \) with \( a_i \leq a_j \), we have \( a_i \leq a_j < a_k \).
   - 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 \( a_i + a_j \leq 2^{k-3} + 2^{k-2} = 3 \cdot 2^{k-3} < a_k \), which is a contradiction.
   - Thus, \( a_i = a_j = a_{k-1} \), so \( a_k = 2a_{k-1} \).

4. **Lemma 3: \( 3 \cdot 2^{k-3} < a_k \leq 2^{k-1} \Rightarrow a_k = 2^{k-1} \):**
   - **Base Case:** For \( k = 1 \), \( a_1 = 1 = 2^0 \).
   - **Inductive Step:** Assume \( a_k = 2^{k-1} \) for some \( k \). We need to show \( a_{k+1} = 2^k \).
     - By Lemma 2, \( a_{k+1} = 2a_k \). But \( a_k = 2^{k-1} \) by the inductive hypothesis.
     - Therefore, \( a_{k+1} = 2 \cdot 2^{k-1} = 2^k \).
     - This completes the inductive step, proving \( a_k = 2^{k-1} \).

5. **Determine the smallest \( n \) such that \( a_n = 100 \):**
   - Since \( a_n = 100 \) and \( a_n \leq 2^{n-1} \) by Lemma 1, we need \( 100 \leq 2^{n-1} \).
   - Solving \( 2^{n-1} \geq 100 \):
     \[
     n-1 \geq \log_2(100) \approx 6.64 \implies n \geq 8
     \]
   - 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.
   - Therefore, \( n \geq 9 \).

6. **Construct the set \( M \) for \( n = 9 \):**
   - The set \( \{1, 2, 4, 8, 16, 32, 64, 96, 100\} \) works for \( n = 9 \).

The final answer is \( \boxed{ \{1, 2, 4, 8, 16, 32, 64, 96, 100\} } \).