Yes. A 6-8-10 triangle for example.

With vertices at $(0,0),(6,0),(0,8)$, the centroid is not a lattice point (but the others are). You need vertices at $(0,0),(36,0),(0,48)$ for it to work.

There are in fact many! A way to see this: 1. Make a triangle whose O is a lattice point (easy). 2. Via translation WLOG O = (0,0). If you triple all coordinates, then G is a lattice point. 3. By the Euler line, doubling all coordinates makes H a lattice point!

@above you must also make the incenter a lattice point

Well that's incredibly annoying. Here are some more general shenanigans to fix that: By scaling it suffices to get all said points to have rational coordinates. 1. Put $I$ literally anywhere with integer coordinates. 2. Draw a circle around $I$ of any rational radius. 3. Pick three rational points on this circle. This is the most restrictive step. To see that there are actually quite a few choices, consider the infinitude of primitive pythagorean triples. 4. The tangent lines at these three points will intersect at rational points, which will be the vertices. 5. Easily we now have H,G,O,I all have rational points. Now go scale.

From the wiki: Quote: The coordinates of the incenter (center of incircle) are $(\dfrac{aA_x+bB_x+cC_x}{a+b+c}, \dfrac{aA_y+bB_y+cC_y}{a+b+c})$, if the coordinates of each vertex are $A(A_x, A_y)$, $B(B_x, B_y)$, and $C(C_x, C_y)$, the side opposite of $A$ has length $a$, the side opposite of $B$ has length $b$, and the side opposite of $C$ has length $c$.

You can also construct this triangle such that these are lattice points: Nine-point center Nagel point Excenters