Let the given points be $A_1, A_2, \ldots, A_{20}$, with $A_i = (x_i, y_i)$, where $x_i, y_i$ are integers.

The centroid of a triangle with vertices $A_i = (x_i, y_i)$, $A_j = (x_j, y_j)$, $A_k = (x_k, y_k)$ is
$$\left(\frac{x_i + x_j + x_k}{3}, \frac{y_i + y_j + y_k}{3}\right).$$

We want to find three points among the $20$ such that both $x_i + x_j + x_k$ and $y_i + y_j + y_k$ are divisible by $3$.

Consider the residues modulo $3$ of the $x$- and $y$-coordinates of each point. Each point can be assigned a pair $(a, b)$, where $a \equiv x_i \pmod{3}$, $b \equiv y_i \pmod{3}$.

There are $3 \times 3 = 9$ possible residue pairs modulo $3$.

By the pigeonhole principle, among $20$ points, at least $\left\lceil \frac{20}{9} \right\rceil = 3$ points must have the same residue pair $(a, b)$. That is, there exist three points $A_i, A_j, A_k$ such that $x_i \equiv x_j \equiv x_k \pmod{3}$ and $y_i \equiv y_j \equiv y_k \pmod{3}$.

Therefore, $x_i + x_j + x_k \equiv 0 \pmod{3}$ and $y_i + y_j + y_k \equiv 0 \pmod{3}$, so the centroid has integer coordinates.

It remains to check that these three points are not collinear. But the problem states that no three of the $20$ points are collinear, so any three points form a triangle.

Thus, there exists a triangle with vertices among the given points whose centroid is also a point with integer coordinates.