First, we will show that for points that are at a distance of $1$ from each other, there are no points $(X, Y)$ that satisfy the condition. Indeed, let us assume that such points exist. Then, $\angle AXB + \angle AYB = 180^\circ$, which means that at least one of these angles is not less than $90^\circ$. Therefore, at least one of the points $X, Y$ must lie inside or on the circle with diameter $AB$, but this circle does not contain any other integer points except for $A$ and $B$.

Now, let's show that for all other pairs of points, such a pair $(X, Y)$ can be found. Let $A = (a_1, a_2)$ and $B = (b_1, b_2)$. If $a_1 \ne b_1$ and $a_2 \ne b_2$, then we can take $X = (a_1, b_2)$ and $Y = (b_1, a_2)$, and $AXBY$

will be a rectangle, which means that it is inscribed. Otherwise, without loss of generality, we can assume that $a_2 = b_2 = t$, and $A = (a_1, t)$ and $B = (b_1, t)$, where $|a_1 - b_1| > 1$. Without loss of generality, we can also assume that $a_1 < b_1$. Then, it is sufficient to take the following points: $X = (a_1 + 1, t - 1)$ and $Y = (a_1 + 1, t + (b_1 - a_1 - 1))$. It is easy to see that the segments $AB$ and $XY$ intersect at the point $K = (a_1 + 1, t)$, and $AK \cdot BK = XK \cdot YK = 1 \cdot (b_1 - a_1 - 1)$, so the quadrilateral $AXBY$ is indeed inscribed.