Suppose $A_1, A_2, \ldots, A_n \subseteq \{1, 2, \ldots, 2018\}$ and $|A_i| = 2$ for $i = 1, 2, \ldots, n$, satisfying that $A_i + A_j$, $1 \leq i \leq j \leq n$, are distinct from each other. Here, $A + B = \{a + b \mid a \in A, b \in B\}$. We aim to determine the maximal value of $n$.

To generalize, let $m = 2018$. We will show that the answer is $2m - 3$ for a general $m$.

Represent $A_i = \{a_1, a_2\}$ with $a_1 < a_2$ by the point $(a_1, a_2)$ in the plane.

Claim: $A_i + A_j = A_i' + A_j'$ if and only if the associated points form a (possibly degenerate) parallelogram with a pair of sides parallel to the line $y = x$.

Proof: Consider the points $(a_1, a_2)$ and $(b_1, b_2)$ in the plane. The sum set $A_i + A_j$ corresponds to the set of sums of coordinates. If $A_i + A_j = A_i' + A_j'$, then the sums must be the same, implying the points form a parallelogram with sides parallel to $y = x$.

Finish: In any right triangle lattice of $m$ points on each of its legs, if there are more than $2m - 1$ vertices chosen, then 4 points will form a parallelogram with a pair of sides parallel to the line $y = x$.

Proof: Let $x_1, \ldots, x_m$ denote the number of points lying on $y = x + c$ for $c = 1, \ldots, m-1$. Consider pairwise differences of points on the same line $y = x + c$. There are $\sum \binom{x_i}{2}$ such differences, and no two can be the same (else a possibly degenerate parallelogram with sides parallel to $y = x$ can be formed). Moreover, each difference must be of the form $r(1, 1)$ for some $r \in [1, m-1] \cap \mathbb{N}$. When $\sum x_i \geq 2m - 2$, we have $\sum \binom{x_i}{2} \geq m$, leading to a contradiction.

For construction, take the $2m - 3$ vertices along the legs of the right triangle.

Thus, the maximal value of $n$ is:
$$\boxed{4033}$$

Note: The original forum solution contained a mistake in the final boxed answer. The correct maximal value of $n$ is $4033$, not $4035$.