We arbitrarily choose n lattice points in a 3-dimensional Euclidean space such that no three points are on the same line. What is the least n in order to guarantee that there must be three points x, y, and z among the n points such that the center of mass of the x-y-z triangle is a lattice point?
The answer $19$ has already been given, including references, so this is just a numerical check of that result.
The specific shape of the lattice is irrelevant, since a point is a lattice point iff it is an integer linear combination of the lattice vectors. The centre of mass of three lattice points is a lattice point iff all coefficients of the sum are multiples of $3$.
To find the maximal number of points without a centre of mass on the lattice, consider the distribution of the residues of the coefficients of the points $\bmod3$ in $\mathbb Z_3^3$. No three points may add up to $0\in\mathbb Z_3^3$.
We cannot have the same point in $\mathbb Z_3^3$ three times. On the other hand, whatever points in $\mathbb Z_3^3$ we do have, we can have them twice, since the only point that would sum to $0$ with the two copies would be a third copy. Thus, we can reduce the problem to finding the maximal number of different points in $\mathbb Z_3^3$ no three of which add up to $0$; then the desired maximal number of not necessarily different points is twice that.
Here's code that enumerates all subsets of $\mathbb Z_3^3$ and finds the maximal size of a subset that doesn't contain a triple that sums to $0$. That maximal size turns out to be $9$, so the maximal size of a set of not necessarily different points is $2\cdot9=18$. Thus one more than that, $19$, is the number of points required to force a set to contain a triple with centre of mass on the lattice.