Colour the squares on the board alternately white and black, like on a chess board. If $n$ is odd, then an $n$-jump always goes to a square of the same colour as the starting square, and if $n$ is even precisely to a square of the other colour. Suppose Mila starts on a white square. If Mila makes a total of $m$ jumps and starts on a white square, Mila ends on a white square if $m$ is of the form $4k$ or $4k + 1$, where $k$ is an integer, and on a black square if $m$ is of the form $4k + 2$ or $4k + 3$.
So the only possibilities for Mila to end up on the initial square are for $m$ of the form $4k$ or $4k + 1$. In the first case, we see immediately that this is indeed possible from the following claim: for every $n$, Mila can return to her starting square after an $(n+1)$-jump, $(n+2)$-jump, $(n+3)$-jump and $(n+4)$-jump. Proof of claim: suppose Mila starts on the square with coordinates $(0,0)$. Then Mila can return by first jumping to $(1, n+1)$, then to $(n+3, n+2)$, then to $(n+4, -1)$ and finally back to $(0,0)$.

For $m = 1$ it is not possible to end on the square where Mila started, but
for $m = 5$ it can be done by jumping as follows:
$$(0, 0) - (1, 1) - (2, 3) - (5, 4) - (1, 5) - (0, 0).$$

Then by pasting the path from the claim for $n = 5, 9, 13, \dots$, we see that
it is thus possible for Mila to end on the square she started, for $m$ of the
form $4k + 1$, as long as $m \ge 5$.
Together we find that Mila can return to her starting square after $m$ jumps
for every $m$ of the form $4k$ or $4k + 1$, for $m \ge 4$ (and thus $k \ge 1$).