Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

Consider sets $A_0,A_1,\dots A_6$. 

Let $A_i=\{x : x\in \mathbb{N},\space 1\leq x\leq 2023, x\equiv i \text{(mod 7)}\}$

Each set has $\frac{2023}{7}=289$ elements.

Form pairs $(x,y)$ such that

$x\in A_1, y\in A_6$

$x\in A_2, y\in A_5$

$x\in A_3, y\in A_4$

As each set has $289$ elements, we can form $289$ pairs from each pair of sets. So, we can form $3\cdot 289=867$ pairs.