$ m = 9,n = 6.$ By Power Mean Inequality we have \[ a + b + c + d\le2\sqrt {1989} < 90. \] Since $ a + b + c + d$ and $ a^2 + b^2 + c^2 + d^2$ have the same parity, the possible values of $ m^2$ are $ 1,9,25,49,81$. But $ m^4 = (a + b + c + d)^2 > a^2 + b^2 + c^2 + d^2 = 1989$, so $ m^2$ is either 49 or 81. Assume that $ m^2 = 49$ and $ \max\{a,b,c,d\} = d$. We have $ a + b + c = 49 - d$ and $ a^2 + b^2 + c^2 = 1989 - d^2$. Since $ (a + b + c)^2 > a^2 + b^2 + c^2$, we have \[ (49 - d)^2 > 1989 - d^2, \] i.e. \[ d^2 - 49d + 206 > 0. \] This gives $ d\ge45$ or $ d\le4$. If $ d\ge45$, then $ d^2\ge2025 > 1989 = a^2 + b^2 + c^2 + d^2$, which is impossible. If $ d\le4$, then $ a + b + c + d\le16$, which is impossible too. Hence $ m^2 = 81$. We have $ n^2\ge\frac {81}4\ge20$ and $ n^2 < \sqrt {1989}\le44$. So $ n^2$ is either 25 or 36. Assume that $ n^2 = 25$. Note that $ 25 - a,25 - b,25 - c$ are all nonnegative. We have $ (25 - a) + (25 - b) + (25 - c) = 19$ and $ (25 - a)^2 + (25 - b)^2 + (25 - c)^2 = 439$. This is impossible since $ 19^2 < 439$. So $ n^2 = 36$. We can see easily that $ m=9,n=6$ is possible. Indeed, $ a=12,b=15,c=18$ satisfies the given equations.

As \[\frac{a^2+b^2+c^2+d^2}{4} > (\frac{a+b+c+d}{4})^2\] By Weighted mean Ineaquality putting values of a+b+c+d = m^2 and that of a^2 + b^2 + c^2 + d^2 =1989 we get m^4 <1989 * 4 m<9.4 as a,b,c,d,are Z+ m=9 max for m=9 a+b+c+d= 81 and a^2 + b^2 + c^2 + d^2 =1989 let max of a,b,c,d is a then b+c+d=81-a and b^2 + c^2 + d^2=1989 -a^2 as (b+c+d)^2 >b^2 + c^2 + d^2 solving this equation we get the value off n as 6