Example 8 Let the set of positive integers be
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{1000}\right\} \text {, }
$$

where, $a_{1}<a_{2}<\cdots<a_{1000} \leqslant 2014$, and
$$
S_{A}=\left\{a_{i}+a_{j} \mid 1 \leqslant i, j \leqslant 1000, i+j \in A\right\}
$$

is a subset of $A$. Find the number of sets $A$ that satisfy the condition. ${ }^{[7]}$
（54th Dutch National Team Selection Exam）
Define a set $A$ that satisfies the conditions as a "good set".
First, prove a lemma.
Lemma: $A$ is a good set if and only if $A = B \cup C$, where
$C \subseteq \{2001, 2002, \cdots, 2014\}$,
$B = \{1, 2, \cdots, 1000 - |C|\}$.
Proof: On the one hand, suppose $A = B \cup C$, where $B$ and $C$ satisfy (1) and (2).
Clearly, $|A| = 1000$, and
$a_i = i (1 \leqslant i \leqslant 1000 - |C|)$.
Consider any $i, j \in \{1, 2, \cdots, 1000\}, i + j \in A$.
At this point, $i + j \leqslant 2000$. Hence, $i + j \in B$.
Let $i = a_t - t$, then $i + t = a_t \in A$.
Thus, $a_i + a_t \in S_A$.
Therefore, $a_i + a_t \in A$.
Notice that, $i = a_t - t \leqslant 1000 - 986 < t$.
Thus, $a_i < a_t \leqslant 1000$.
Then, $a_t < a_i + a_t \leqslant 2000 < a_{t+1}$, which contradicts $a_i + a_t \in A$.
Therefore, $a_t = t$. Hence, $B = \{1, 2, \cdots, t\}$.
This shows that $A = B \cup C$, where $B$ and $C$ satisfy (1) and (2).
The lemma is proved.
According to the lemma, a good set $A = B \cup C$ is uniquely determined by the choice of $C$ in (1) and (2).

Since the number of ways to choose $C$ is $2^{14}$, the number of good sets is $2^{14}$.

[Note] The core task of this problem is to characterize the intrinsic structure of good sets. In comparison, the final calculation of the number of good sets is extremely trivial.