Solution:

Suppose we have an odd number of arbitrary points $A_1, A_2, \ldots, A_{2k + 1}$ then we claim that if we take the vector $A_iA_j$ for $i < j$ and $j - i$ odd and the vector $A_jA_i$ for $i < j$ and $j - i$ even, then we get the sum of the vectors zero. We prove the claim by induction. It is true for $k = 3$ because we have $A_1A_2 + A_2A_3 + A_3A_1 = 0$. So suppose it is true for $2k - 1$. The additional vectors when we move to $2k + 1$ are $A_{2k}A_{2k + 1}$, $\sum A_{2i + 1}A_{2k}$, $\sum A_{2i}A_{2k + 1}$, $\sum A_{2k}A_{2i}$ and $\sum A_{2k + 1}A_{2i + 1}$. But $A_{2i + 1}A_{2k} + A_{2k}A_{2i} + A_{2i}A_{2k + 1} + A_{2k + 1}A_{2i + 1} = A_{2i + 1}A_{2i} + A_{2i}A_{2i + 1} = 0$, leaving the three terms $A_{2k + 1}A_1$, $A_1A_{2k}$ and $A_{2k}A_{2k + 1}$ which also sum to zero. Hence the result is true for $2k + 1$ and hence for all odd numbers. Thus for $mn$ odd we can number the centers of the black squares in an arbitrary fashion, use the rule given for the arrow directions and then the vectors for the black squares will sum to zero. Similarly for the white squares.

However, the same general result is not true for an even number of points. So we need something else for $mn$ even. Let $B$ be the center of the black square at one end of the main diagonal and $W$ be the center of the white square at the other end. Let $B_1, B_2, \ldots, B_{2k + 1}$ be the centers of the other black squares and $W_1, W_2, \ldots, W_{2k + 1}$ the centers of the other white squares. Take vectors $BB_i$, $WW_i$ and for $B_iB_j$ and $W_iW_j$ take the same rule as before (if $i < j$ take $B_iB_j$ if $j - i$ is odd and $B_jB_i$ if $j - i$ is even, similarly for $W_iW_j$). Now consider square centers $X$, $Y$ which are symmetric with respect to the center of the rectangle (in other words the center of the rectangle is the midpoint of $XY$). The pairs $(X, Y)$ are of three types: opposite colors, both white and both black. The number of both white pairs must equal the number of both black pairs since the number of white and black squares is equal. For the first type we have $BX + WY = 0$. For the second type we have $WX + WY = WB$ and for the third type we have $BX + BY = BW$. Since they are equal in number the second and third type sum to zero.