Denote $S$ as a subset of $\{ 1, 2, 3, \ldots, 1000 \}$ with the property that no sum of two different elements in $S$ is itself an element of $S$. We wish to find the maximum number of elements in $S$.

To address this problem, consider the possibility of selecting elements from $\{ 1, 2, 3, \ldots, 1000 \}$ such that adding any two distinct elements never results in another element within the same set. An effective approach to create such a subset involves selecting alternating numbers starting from $1$.

Construct $S$ by including every second integer from the set:
$$S = \{ 1, 3, 5, 7, \ldots, 999 \}$$

To verify that this set satisfies the condition:
- Let $a, b \in S$ with $a < b$. The sum $a + b$ will be:
- At least $1 + 3 = 4$,
- At most $999 + 997 = 1996$.

Notice that every element $a$ in $S$ is odd and any sum $a+b$ of two distinct odd numbers is even. Therefore, no sum of two elements in the constructed set forms another odd number in the sequence. Thus, $a + b \notin S$.

Since there are 500 odd numbers in $\{1, 2, 3, \ldots, 999\}$, and by the inclusion of 1 as the first odd number, there are 501 numbers in our subset $S$.

Therefore, the maximum number of such elements in $S$ is:
$$\boxed{501}$$