In a $n$ by $n$ chessboard, the rook can move along each row and then move one square to the next row and go down that row and then move one more square to the next row, etc... until the rook has covered the whole board. For each trip down a row of the board and then taking the extra move to the next row, there are $n-1+1=n$ moves. Since there are $n$ columns and the last row does not have an extra move, we have that a there are a total of $n^2-1$ moves required to cover the whole board. From there, there are either $n-1$ or $2(n-1)$ moves required to get back to the original corner. Thus, solving the equations $n^2-1+n-1=n^2$ and $n^2-1+2n-2=n^2$ yields $n=\boxed{1,2}$

For odd $n>1$, the answer is no. When the rook visits a square in one particular row, it must immediately visit another square in the same row, and then leave; so the squares in a given row must be visited in pairs. If the rook tries to complete an initially empty row with an odd number of squares, it will get stuck. For even $n$, the tour is possible. Suppose we start in the bottom-left corner, $(n,1)$. First we move to $(1,1)$ and visit all the squares in the first two rows using the single-step pattern right-down-right-up, ending with a downward move into $(2,n)$ before going back to $(2,1)$. Then we move to $(3,1)$ and repeat the pattern, two rows down. Repeating this for every pair of rows lands us in $(n,n)$, from which we can return to $(n,1)$. [asy][asy] int n=8;draw((0,n)--(n,n)--(n,0)); for(int i=0;i<n;++i){draw((0,i)--(n,i));draw((i,0)--(i,n));} for(int i=0;i<n;++i)for(int j=0;j<n;++j)dot((i+.5,j+.5)); currentpen=magenta+linewidth(.4mm); draw((.5,.5)--(.25,.75)--(.25,n-.75)--(.5,n-.5)); path p=(0,0)--(1,0)--(1,1)--(2,1)--(2,0); for(int i=0;i<n;i=i+2){for(int j=2;j<n;j=j+2){draw(shift((j-.5,i+.5))*p);}draw((.5,i+1.5)--(1.5,i+1.5)--(1.5,i+.5)^^(n-.5,i+.5)--(n-.75,i+.25)--(.75,i+.25)--(.5,i+.5));if(i>0){draw((.5,i+.5)--(.5,i-.5));}} [/asy][/asy]