Let's consider the set $S = \{1, 2, 3, \ldots, 3000\}$. We are tasked with determining whether there exists a subset $A \subseteq S$ such that $|A| = 2000$ and for every element $x \in A$, the element $2x \notin A$.

To solve this problem, we first think about the constraints imposed by the requirement $x \in A \implies 2x \notin A$. This means that for each $x$ included in the subset $A$, its double $2x$ cannot also be in $A$.

1. Counting the Valid Elements:
- A number $x$ in set $S$ can have its double $2x$ in the same set provided $2x \leq 3000$.
- The largest integer $x$ for which $2x \leq 3000$ is $x = 1500$.

2. Construction of Subset:
- We can construct $A$ such that $A$ contains as many numbers as possible, none of which have their doubles also included.
- We start by considering the largest possible numbers in $S$ which do not have their doubles in $S$. These are $1501, 1502, \ldots, 3000$.

3. Counting the Available Space:
- The subset $\{1501, 1502, \ldots, 3000\}$ consists of $3000 - 1500 = 1500$ elements.
- This subset satisfies the condition because the smallest element $1501$ when doubled is greater than 3000, hence not in $S$.

4. **Attempt to Extend $A$:**
- We need at least 2000 elements in $A$. The previous subset only has 1500 elements.
- To achieve 2000 elements, any additional elements must come from the set $\{1, 2, \ldots, 1500\}$ without violating the condition $2x \notin A$.

5. Limitations:
- Selecting any value from $\{1, 2, \ldots, 1500\}$ means its double $2x$ also has the potential to be in $\{1, 2, \ldots, 3000\}$, specifically in $\{2, 4, \ldots, 3000\}$.
- This constraint makes it impossible to construct a valid set $A$ of size 2000, as any expansion beyond the initial 1500 would immediately conflict with the condition.

Hence, the answer to the question is:

$$\boxed{\text{No}}$$
This solution demonstrates that it is not possible to construct a subset $A \subseteq S$ that contains 2000 elements while satisfying the condition that no element has its double also in the subset.