Question 770026
One of the distances must be odd and the other distance must be even.


Let m and n be natural numbers.
Distances of the sides of the rectangle are 2m and 2n+1.  We have no restriction on the order between m and n.  


The sum of the distances of each direction on this rectangle (the length plus width) is 2m+2n+1=2(m+n)+1;This sum is an odd number.
The perimeter will be 2*(2(m+n)+1)=4(m+n)+2, which is even.
  No further result on the sum is possible without having more information.