A subset $M$ of $\{1, 2, . . . , 2006\}$ has the property that for any three elements $x, y, z$ of $M$ with $x < y < z$, $x+ y$ does not divide $z$. Determine the largest possible size of $M$.
1. **Restate the problem with a stronger condition:**
   Consider the stronger condition that there does not exist distinct elements \( x, y, z \in M \) such that \( x + y = z \). Suppose \( M = \{a_1, a_2, \ldots, a_k\} \) is a subset of such elements.

2. **Transform the set \( M \):**
   Define \( M' = \{a_2 - a_1, a_3 - a_1, \ldots, a_k - a_1\} \). This transformation shifts all elements of \( M \) by subtracting \( a_1 \) from each element except \( a_1 \) itself.

3. **Disjoint sets condition:**
   We must have \( M \setminus \{a_1\} \) and \( M' \) disjoint. If they were not disjoint, there would exist \( a_i \in M \setminus \{a_1\} \) and \( a_j - a_1 \in M' \) such that \( a_i = a_j - a_1 \). This implies \( a_i + a_1 = a_j \), which contradicts the condition that \( x + y \neq z \) for any \( x, y, z \in M \).

4. **Cardinality of the union of disjoint sets:**
   Since \( M \setminus \{a_1\} \) and \( M' \) are disjoint, we have:
   \[
   |M \setminus \{a_1\}| + |M'| = |M \setminus \{a_1\} \cup M'|
   \]
   The size of \( M \setminus \{a_1\} \) is \( k-1 \), and the size of \( M' \) is also \( k-1 \). Therefore:
   \[
   |M \setminus \{a_1\} \cup M'| = 2(k-1)
   \]

5. **Bound on the size of the union:**
   Since \( M \setminus \{a_1\} \cup M' \) is a subset of \(\{1, 2, \ldots, 2006\}\), we have:
   \[
   2(k-1) \leq 2006
   \]
   Solving for \( k \):
   \[
   2(k-1) \leq 2006 \implies k-1 \leq 1003 \implies k \leq 1004
   \]

6. **Construct a valid set \( M \) with 1004 elements:**
   Consider the set \( M = \{1003, 1004, \ldots, 2006\} \). This set has 1004 elements. We need to verify that it satisfies the condition that for any three elements \( x, y, z \in M \) with \( x < y < z \), \( x + y \) does not divide \( z \).

   - For any \( x, y \in M \), \( x + y \geq 1003 + 1004 = 2007 \).
   - Since the maximum element in \( M \) is 2006, \( x + y \) is always greater than any element \( z \in M \).

   Therefore, the set \( M = \{1003, 1004, \ldots, 2006\} \) satisfies the given condition.

The final answer is \( \boxed{ 1004 } \)