The answer is $1004$.

When $M = \{1003, 1004, \dots, 2006\}$, the sum of any two elements in $M$ is at least $2007$, which is larger than the largest element in $M$. Therefore, $x + y$ does not divide $z$ for any $x, y, z \in M$. This gives a possible case for $|M| = 1004$.

Next, consider any subset $M$ satisfying the constraint. Suppose the largest element of $M$ is $a$. Then $M$ contains at most one element in each of the pairs
$$(1, a-1), (2, a-2), \dots, \left( \left\lfloor \frac{a}{2} \right\rfloor, \left\lceil \frac{a}{2} \right\rceil \right)$$
(note that the last pair may contain the same number, which does not affect the validity of our claim). Therefore, $M$ contains at most $\frac{a}{2}$ elements smaller than $a$. Since $a$ is the largest element, $M$ cannot contain elements larger than $a$. Therefore,
$$|M| \le \frac{a}{2} + 1 \le 1003 + 1 = 1004.$$
Hence, the largest possible size of $M$ is $1004$.