Loosely inspired by the game [Abalone](https://en.wikipedia.org/wiki/Abalone_(board_game)), I've encountered the following simple problem I cannot solve.  

Suppose that we are given a finite set of marbles on an infinite chessboard.  

One move consists of one marble jumping over another to an empty space.  

For example, if (0,0) and (1,0) have marbles, but (2,0) doesn't, then we can move (0,0) to (2,0).  

It is easy to see that there are starting configurations of an even number of marbles that we can "march off" to infinity using such moves.  

But is this possible to do with an odd number of marbles?  

It feels like this should have a simple explanation, but I don't see any right now.


More on motivation: In Abalone it is useful to start by moving to the center of the board. If one doesn't make sideways moves, then the triangular Abalone board reduces to a chess board. Or almost, because in Abalone it is also allowed to move adjacent marbles in any direction, not just in the direction of the row they are contained in. If one allows that too, then there are more configurations that can do an infinite march. Also, in Abalone one can move at most 3 marbles, so to move as fast as possible, one would always move 3 marbles, and not 2 at a time, but whatever argument works for parity, probably also works for divisibility with 3.



Can an odd number of marbles jump to infinity?
With $5$ you can using the following moves:



```
.....  .....  .....  .....  .....  .....  .....  ..oo.  ...oo 
.....  ..o..  ..o..  ..o..  ..oo.  ...oo  ..ooo  ..ooo  ..ooo 
.oo..  .oo..  .oo..  ..oo.  ..oo.  ..oo.  ..oo.  .....  ..... 
ooo..  oo...  .oo..  ..oo.  ..o..  ..o..  .....  .....  ..... 

```

So the only numbers of pieces for which no configuration can go to infinity are $1$ and $3$.


