$m = 9,n = 6.$ 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