4. Let $A_{1}, A_{2}, \cdots, A_{n}$ be two-element subsets of the set $\{1,2, \cdots, 2018\}$, such that the sets $A_{i}+A_{j}(1 \leqslant i \leqslant j \leqslant n)$ are all distinct, where,
$$
A+B=\{a+b \mid a \in A, b \in B\} .
$$

Find the maximum possible value of $n$.
4. The maximum value of $n$ is 4033.

If there exists $A_{i}=A_{j}(i \neq j)$, then $A_{i}+A_{i}=A_{j}+A_{j}$, which contradicts the condition.
Thus, $A_{1}, A_{2}, \cdots, A_{n}$ are all distinct.
For a binary set $A=\{a, b\}(a<b)$, let $S_{i}=\{x-y \mid x, y \in M_{i}, x>y\}$.
If there exists $i \neq j$ such that $S_{i}$ and $S_{j}$ intersect, then there exist $x, y \in M_{i}, u, v \in M_{j}, x>y, u>v$, satisfying
$$
x-y=u-v \Rightarrow x+v=y+u.
$$

Thus, in $A_{1}, A_{2}, \cdots, A_{n}$, there are four sets
$$
\left\{x, x+d_{i}\right\}, \left\{y, y+d_{i}\right\}, \left\{u, u+d_{j}\right\}, \left\{v, v+d_{j}\right\},
$$

satisfying
$$
\left\{x, x+d_{i}\right\}+\left\{v, v+d_{j}\right\}=\left\{y, y+d_{i}\right\}+\left\{u, u+d_{j}\right\},
$$

which contradicts the condition. Hence, $S_{1}, S_{2}, \cdots, S_{m}$ are mutually disjoint.
Since the elements in set $M_{i}$ are all no more than 2017, and the elements in set $S_{i}$ are all positive integers no more than 2016, then
$$
\sum_{i=1}^{m}\left|S_{i}\right| \leqslant 2016.
$$

Let $M_{i}=\left\{x_{1}, x_{2}, \cdots, x_{n_{i}}\right\}\left(x_{1}<x_{2}<\cdots<x_{n_{i}}\right)$. Then set $S_{i}$ contains at least $n_{i}-1$ distinct numbers
$$
x_{2}-x_{1}<x_{3}-x_{1}<\cdots<x_{n_{i}}-x_{1}.
$$

Thus, $\left|S_{i}\right| \geqslant n_{i}-1(1 \leqslant i \leqslant m)$.
Combining (1) and (2), and noting that $d\left(A_{i}\right) \leqslant 2017$, we have $m \leqslant 2017$.
Therefore, $2016 \geqslant \sum_{i=1}^{m}\left|S_{i}\right| \geqslant \sum_{i=1}^{m}\left(n_{i}-1\right)=n-m$
$$
\Rightarrow n \leqslant m+2016 \leqslant 2017+2016=4033.
$$

For $1 \leqslant i \leqslant 2017$, let $A_{i}=\{1, i+1\}$; for $2018 \leqslant j \leqslant 4033$, let $A_{j}=\{j-2016,2018\}$.

We now verify that the 4033 sets $A_{1}, A_{2}, \cdots, A_{4033}$ satisfy the requirements.

Note that, the sets with distances $1,2, \cdots, 2016$ each have 2 sets, and the set with distance 2017 has 1 set,
$$
\begin{array}{l}
M_{1}=\{1,2017\}, M_{2}=\{1,2016\}, \cdots, \\
M_{2016}=\{1,2\}, M_{2017}=\{1\}.
\end{array}
$$

For a finite set $A \subset \mathbf{Z}$, translating $A$ so that the minimum element is 0, i.e., $A-\min A$, is called the "standardization" of $A$.

For two binary sets $A, B$ with distance $d$, the standardization of $A+B$ is
$$
\{0, d\}+\{0, d\}=\{0, d, 2 d\},
$$

i.e., $A+B$ is a ternary set and is an arithmetic sequence with common difference $d$;
For two binary sets $A, B$ with distances $d_{1}, d_{2}$, $d_{1} \neq d_{2}$, the standardization of $A+B$ is
$$
\left\{0, d_{1}\right\}+\left\{0, d_{2}\right\}=\left\{0, d_{1}, d_{2}, d_{1}+d_{2}\right\},
$$
$A+B$ is a quaternary set, and the standardization of $A+B$ is uniquely determined by $d_{1}, d_{2}$.

Therefore, to verify that $A_{i}+A_{j}(1 \leqslant i \leqslant j \leqslant 4033)$ are all distinct, it suffices to verify for each $1 \leqslant i \leqslant 2017$ that $x+y$ $\left(x, y \in M_{i}, x \leqslant y\right)$ are all distinct, and for each $1 \leqslant i < j \leqslant 2017$, that $x+y\left(x \in M_{i}, y \in M_{j}\right)$ are all distinct. Direct verification shows that this is true.