Does the contrapositive permits to conclude ...
so I consider your question from the point of view of logic.
Consider the following propositions:
$P_1=$ The set of masses under consideration is finite.
$P_2=$ The set of masses under consideration is non-empty.
$Q=$ The system has a gravity center.
You claim that $P1 \wedge P2\to Q$.
This is not necessarily true, as shown by Ben Crowell. You may add some assumptions, especially if you want the center of mass be calculated by the formula you wrote. For instance that the space is Euclidean, so we are talking about Newtonian mechanics, or that we are talking about special relativity, provided that the masses are considered at rest w.r.t. a given inertial reference frame. Let the proposition containing these assumptions be $P_3$.
Now, your conditional statement is
$$P1 \wedge P2 \wedge P3\to Q.$$
The contraposition is
$$\neg Q \to \neg (P1 \wedge P2 \wedge P3).$$
$$\neg Q \to \neg P1 \vee \neg P2 \vee \neg P3.$$
Hence, if the system doesn't have a gravity center, the contraposition only says that at least one of the propositions $P_1$, $P_2$, and $P_3$ is false.
It can be that $P_1$ is false, hence the set of masses is infinite. For instance, the space is Euclidean, and there are equal masses at the vertices of a cubic lattice extending in all the space. This set is countable. Or, there is a countable set of masses, which increase in a divergent way, such as, $m_i=i\cdot m_0$, where $m_0>0$.
Another example when a countable set of masses don't have a center of mass is the one you gave
Ex. : within a 2D infinite plan, an infinite set of equal masses
linearly distributed along the x and y axis will have a gravity center
at its origin O.
This set doesn't actually have a gravity center. You say it is at the origin, but there is no absolute origin in the Euclidean plane. If you change the origin, you will find the gravity center at another point.
But merely the negation of $P_1$ doesn't imply that there is no center of gravity, and Emilio Pisanty gave such an example from continuous mechanics.
Another way for $P1 \wedge P2 \wedge P3$ to be false is that $P_2$ is false, hence there are no masses. It follows equally from contraposition.
Or, if both $P_1$ and $P_2$ are true, then $P_3$ is false, for instance the theory is not Newtonian mechanics or special relativity. Ben Crowell exemplified with general relativity.
The main point is that, if your conditional statement is true, contraposition only implies that at least one of the conditions in your statement is false, and it doesn't allow you to pick which one you want to invalidate.