One, (20 points) For the four-digit number $\overline{a b c d}$, the sum of its digits $a + b + c + d$ is a perfect square. Reversing the digits forms the four-digit number $\overline{d c b a}$, which is 4995 greater than the original number. Find all such four-digit numbers.
$-, \Pi(1000 d+100 c+10 b+a)-(1000 a+100 b$ $+10 c+d)=4995$ simplifies to
$$
111(d-a)+10(c-b)=555 \Rightarrow 5 \mid(d-a) .
$$

If $d-a$ is $0, -5$, the equation cannot hold, so it must be $d-a=5 \cdot 11 c-b=0$.
$$
\begin{array}{l}
\text { Given } a+b+c+d=n^{2}, \text { then } \\
a+b+b+a+5=n^{2} .
\end{array}
$$

Thus $2(a+b)=n^{2}-5 \Rightarrow n$ is an odd number 1. $n^{2}>5$.
$$
\begin{array}{l}
\because a+b+c+d<4 \times 9=36 . \\
\therefore n^{2}<36 .
\end{array}
$$

So $n=3$ or 5.
When $n=3$, $a+b=2$. Since $a \neq 0$, there are two cases:
T $a=2, b=0$. In this case $c=b=0, d=a+5=7$.
(2) $a=1, b=1$, in this case $c=1, d=6$.

When $n=5$, $a+b=10$. There are four cases:
(1) $a=1, b=9$, in this case $c=9, d=6$.
(2) $a=2, b=8$, in this case $c=8, d=7$.
(3) $a=3, b=7$, in this case $c=7, d=8$.
(4) $\dot{a}=4, b=6$, in this case $c=6, d=9$.

Therefore, the required four-digit numbers are $2007,1116,1996,2887,3778$, 4669.