I think I find the answer $mn$ should be even

first we change question without calculate square itself so the number of unit square which are painted the same color as that square and which have at least one common vertex with it is odd if $m$ and $n$ are both odd then if we sum all this number it is e odd number (because we have odd unit square) but it is obvious that this sum should be even so at least one of $m$ or $n$ should even . for coloring : WLOG $m$ is even so we have $m$ row $A_{1},...a_{m}$ $A_{1}$ and $A_{2}$ all unit square black $A_{3}$ and $A_{4}$ all unit square white $A_{5}$ and $A_{6}$ all unit square black ...