Three. (50 points) Let the set of positive integers be
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{1000}\right\},
$$

where, $a_{1}<a_{2}<\cdots<a_{1000} \leqslant 2017$. If for any $1 \leqslant i, j \leqslant 1000, i+j \in A$ there is $a_{i}+a_{j} \in A$, find the number of sets $A$ that satisfy the condition.
Three, let $k$ be the largest positive integer such that $a_{k} \leqslant 2000$. Then
$$
a_{1000}>a_{999}>\cdots>a_{k+1} \geqslant 2001 .
$$

If $a_{k}>1000$, then there exists a positive integer $i$, such that
$$
a_{k}=1000+i \in A \text {. }
$$

Thus, $a_{1000}+a_{i} \in A$.
But $a_{1000}+a_{i}>a_{1000}$, which is a contradiction.
Therefore, $a_{i} \leqslant 1000(i=1,2, \cdots, k)$.
Notice that, $a_{i} \geqslant i(i=1,2, \cdots, 1000)$.
Hence $a_{k} \geqslant k$.
If $a_{k}>k$, then there exists a positive integer $i$, such that $i=a_{k}-k$, i.e., $i+k=a_{k} \in A$.
Thus, there exists a positive integer $m$, such that
$$
a_{m}=a_{i}+a_{k} \in A \text {. }
$$

Also, $i+k=a_{k} \leqslant 1000$, and the number of elements in $A$ that are not less than 2001 is at most 17, thus,
$$
\begin{array}{l}
k \geqslant 1000-17=983, \\
i \leqslant a_{k}-k \leqslant 1000-k \leqslant 17a_{m}=a_{i}+a_{k}>a_{k}$, which contradicts the maximality of $a_{k}$.
Therefore, $a_{k}=k$.
Hence $A=\left\{1,2, \cdots, k, a_{k+1}, a_{k+2}, \cdots, a_{1000}\right\}$, where $a_{k+1}, a_{k+2}, \cdots, a_{1000}$ are all greater than 2000.

When $k=983,984, \cdots, 1000$, the number of sets $A$ that satisfy the conditions is
$$
\mathrm{C}_{17}^{17}+\mathrm{C}_{17}^{16}+\cdots+\mathrm{C}_{17}^{1}+\mathrm{C}_{17}^{0}=2^{17} .
$$

Therefore, the number of sets $A$ that satisfy the conditions is $2^{17}$.