Test $\mathbf{E}$ Let $a, b, c, d, m, n$ be positive integers,
$$
a^{2}+b^{2}+c^{2}+d^{2}=1989, a+b+c+d=m^{2} \text {, }
$$

and the largest of $a, b, c, d$ is $n^{2}$, determine (and prove) the values of $m, n$.
By Cauchy inequality,
$$
\begin{aligned}
(a+b+c+d)^{2} & \leqslant(1+1+1+1)\left(a^{2}+b^{2}+c^{2}+d^{2}\right) \\
& \Rightarrow m^{2} \leqslant 2 \sqrt{1989}a^{2}+b^{2}+c^{2}+d^{2} \Rightarrow m^{2}=49,81.
\end{aligned}
$$
Assume without loss of generality that $a \leqslant b \leqslant c \leqslant d=n^{2}$. If $m=49$, then
$$
\begin{aligned}
(49-d)^{2} & =(a+b+c)^{2}>a^{2}+b^{2}+c^{2}=1989-d^{2} \\
& \Rightarrow d^{2}-49 d+206>0 \Rightarrow d>44 \text { or } d \leqslant 4.
\end{aligned}
$$

However,
$$
\begin{array}{c}
45^{2}>1989>d^{2} \Rightarrow d<45,
\end{array}
$$

so $m^{2} \neq 49$. Therefore, $m^{2}=81 \Rightarrow m=9, d=n^{2} \in\{25,36\}$.
If $d=n^{2}=25$, let
$$
a=25-p, b=25-q, c=25-r(p, q, r \geqslant 0),
$$

then by the given conditions we have
$$
p+q+r=19, p^{2}+q^{2}+r^{2}=439,
$$

which contradicts $(p+q+r)^{2}>p+q+r$. Therefore, $n^{2}=36$, i.e., $n=6$. From this, it is not difficult to deduce that $a=12, b=15, c=18, d=36$.