a)If the rook returns to it's initial place, then all the moves which were made in a direction must be made in the opposite direction hence the sum of all the moves must be even, however $S=\frac{2013\cdot 2014}{2}=2013\cdot 1007$ which is obviously odd, therefore it isn't possible. b) Claim 1 If the number $n$ is divisible by $4$, then there is an order of moves such that the condition in the problem is met Proof: We divide it in $\frac{n}{4}$ groups of moves. Then every group will be of the form $4k, 4k+1, 4k+2, 4k+3$. Now, we move $4k$ moves horizontally on the right of the initial square, then we move in order, $4k+1$ and $4k+2$ moves horizontally on the left, then we move the rook $4k+3$ horizontally on the right and we arrive in the initial square. Claim 2 If the number $n-3$ is divisible by $4$, then there is an order of moves such that the condition in the problem is met Proof: The first 3 moves we will just move the rook $1$ and $2$ squares on the right, and then $3$ squares on the left and we arrive to our initial square. Now we just use Claim 1 and the prof is complete. Also, if 4 divides $n-1$ and $n-2$ then the sum of all the moves made in the end is an odd number but this contradicts our initial statement in (a). Hence $n$ can be $\equiv 3,0 (mod4)$ and the statement will follow. Also, in the original problem point b was how many numbers in the following group $\{ 1,2 \cdots 2012\}$ have that property. For this question, the answer is $1006$.