jmc

Since the perimeter is 142, the sides of the rectangle add up to $142/2=71.$ Let $x$ be one side length of the rectangle. Then the other side length is $71-x,$ so the area is $$x(71-x)=71x-x^2.$$Completing the square, we get $$-x^2+71x=-x^2+71x-\frac{71^2}{2^2}+\frac{71^2}{2^2}=\frac{5041}{4}-{\left(x-\frac{71}{2}\right)}^2.$$To minimize this, we want $x$ as close as possible to $\frac{71}{2}.$ Normally, we could take $x=\frac{71}{2},$ but $x$ must be an integer, so we can take $x$ as either 35 or 36.

Thus, the maximum area of the rectangle is $35\cdot 36=\boxed{1260}.$