Claim: the only such numbers are the multiples of $3$. If $k=3t$ pick an axis and grid aligned $k\times k$ square divided in $t\times t$ square each of side $3\times 3$, and divide these in two triangles. In this way it is straight forward enough to see that the barycenters must have integer coordinates. Lemma: given triangles with integer centroids (calls such triangles good) $ABC$ and $A'BC$, the vertices $A$ and $A'$ differ by three times an integer vector Proof: $A'-A=3(\frac{A'+B+C}{3}-\frac{A+B+C}{3})$. Since the two centroids have integer coordinates, their differences is an integer vector, so we are done. Now, using the lemma, divide the vertices of the triangulation in ($\leq$) $3$ equivalence classes where the equivalence relation is differing by three times an integer vector [from any vertex we can reach a vertex of a fixed triangle, by multiple steps of the form $A\to A'$ if there is a segment $BC$ such that $ABC,A'BC$ both belong to the triangulation]. Since the square has $4>3$ vertices, at least two will be in the same of the three equivalence classes. Let $\Delta x$ be the difference between the $x$ coordinates of these two vertices, and $\Delta y$ that of $y$ coordinates. If the two vertices are opposite wrt the square we have $\Delta x^2+\Delta y^2=2k^2$. Since the two vertices are equivalent, $3|\Delta x,\Delta y\implies 3|\Delta x^2+\Delta y^2=2k^2\implies 3|k$. Similarly if the two vertices are adjacent wrt the square, $3|\Delta x^2+\Delta y^2=k^2\implies 3|k$. In both cases, $k$ is divisible by $3$, so we are done.