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

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,
$$
\begin{array}{l}
C \subseteq \{2001, 2002, \cdots, 2014\}, \\
B = \{1, 2, \cdots, 1000 - |C|\}.
\end{array}
$$

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 $i, j \in \{1, 2, \cdots, 1000\}$, such that $i + j \in A$.
At this point, $i + j \leqslant 2000$. Hence, $i + j \in B$.
Then, $i, j \in B$. 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 $2^{14}$.