Is there a triangle whose vertices, as well as the four classical points, the centroid, the orthocenter, the incenter, and the circumcenter, all have integer coordinates?

Clearly it is enough to find a triangle in which all seven points are rational, as then you can make them integral by rescaling. But given any triangle with rational coordinates, aren't the centroid, orthocenter, and circumcenter all automatically at rational coordinates?
So the only nonautomatic point is the incenter. If $A$, $B$, and $C$ are the coordinates of the corners, and $a$, $b$, and $c$ are the side lengths of the respective opposite sides (so that $a = \BC\$, for example) then the incenter is at $(aA + bB + cC) / (a + b + c)$. This is not necessarily rational — the triangle with coordinates $(0,0)$, $(1,0)$, and $(0,1)$ is a counterexample — but is rational as soon as the three sidelengths are. So Noam D. Elkies's answer in the comments has many generalizations, including all Pythagorean triples, and also including combinations of them. 

