Romanian Mathematical Olympiad - Final Round

We will attach integer coordinates to each square of the board, the starting point being $(0,0)$. Then, after each move one coordinate will be of the form $\pm 1 \pm 3 \pm 5 \pm \dots$, and the other will be of the form $\pm 2 \pm 4 \pm 6 \pm \dots$.

a) If $(0,0)$ could be reached in exactly $2013$ moves, then it would be possible to choose the $+$ and $-$ signs so that $\pm 1 \pm 3 \pm 5 \pm \dots \pm 2013 = 0$ and $\pm 2 \pm 4 \pm 6 \pm \dots \pm 2012 = 0$.
But, for each combination of signs, $\pm 1 \pm 3 \pm \dots \pm 2013$ has the same parity as $1+3+\dots+2013 = 1007^2$ is odd, so the equality $\pm 1 \pm 3 \pm 5 \pm \dots \pm 2013 = 0$ is impossible. This shows that $(0,0)$ cannot be reached in $2013$ moves.

b) If $(0,0)$ can be reached in $n$ moves, then
$$\pm 1 \pm 3 \pm 5 \pm \dots \pm \left( 2 \lfloor \frac{n+1}{2} \rfloor - 1 \right) = 0 \quad \text{and} \quad \pm 2 \pm 4 \pm 6 \pm \dots \pm 2 \lfloor \frac{n}{2} \rfloor = 0$$
Relation above implies that $0$ has the same parity as $1+3+\dots+(2\lfloor\frac{n+1}{2}\rfloor-1) = \lfloor\frac{n+1}{2}\rfloor^2$, so $\lfloor\frac{n+1}{2}\rfloor$ must be even, that is $4 \mid n$ or $4 \mid n+1$.
From the second relation, $0$ has the same parity as $1+2+\dots+\lfloor\frac{n}{2}\rfloor = \frac{\lfloor\frac{n}{2}\rfloor(\lfloor\frac{n}{2}\rfloor+1)}{2}$, that is $4 \mid \lfloor\frac{n}{2}\rfloor$ or $4 \mid \lfloor\frac{n}{2}\rfloor+1$. This shows that $8$ must divide one of the numbers $n-1, n, n+1$ or $n+2$.
From these, it follows that $n$ must be of the form $8k$ or $8k-1$, $k \in \mathbb{N}^*$.
For each $k \in \mathbb{N}^*$, $(0,0)$ can be reached in $8k$ moves, because $(1-3-5+7)+ (9-11-13+15)+\dots+((8k-7)-(8k-5)-(8k-3)+(8k-1)) = 0$ and $(2-4-6+8)+(10-12-14+16)+\dots+((8k-6)-(8k-4)-(8k-2)+8k) = 0$.

It is also possible to reach $(0,0)$ in $8k - 1$ moves, because $(1 - 3 - 5 + 7) + (9 - 11 - 13 + 15) + \dots + ((8k - 7) - (8k - 5) - (8k - 3) + (8k - 1)) = 0$ and $(2+4-6) + (8-10-12+14) + \dots + ((8k-8) - (8k-6) - (8k-4) + (8k-2)) = 0$.
Therefore, the answer is: all $n = 8k$ or $n = 8k - 1$, $k \in \mathbb{N}^*$.