Let $\overline{abcd}$ be one of the 9999 numbers $0001, 0002, 0003, \ldots, 9998, 9999$. Let $\overline{abcd}$ be an [i]special[/i] number if $ab-cd$ and $ab+cd$ are perfect squares, $ab-cd$ divides $ab+cd$ and also $ab+cd$ divides $abcd$. For example 2016 is special. Find all the $\overline{abcd}$ special numbers.
[b]Note:[/b] If $\overline{abcd}=0206$, then $ab=02$ and $cd=06$.
1. Let $\overline{abcd}$ be a special number. This means that $ab - cd$ and $ab + cd$ are perfect squares, $ab - cd$ divides $ab + cd$, and $ab + cd$ divides $abcd$.
2. Let $ab = A$ and $cd = B$. Then, we have:
   \[
   A - B = x^2 \quad \text{and} \quad A + B = y^2
   \]
   for some integers $x$ and $y$.
3. Since $A - B$ divides $A + B$, we have:
   \[
   x^2 \mid y^2
   \]
   This implies that $y^2 = kx^2$ for some integer $k$.
4. Since $A + B$ divides $100A + B$, we have:
   \[
   y^2 \mid 100A + B
   \]
   Substituting $A + B = y^2$ and $A - B = x^2$, we get:
   \[
   y^2 \mid 100A + B \implies kx^2 \mid 100A + B
   \]
5. From $A + B = y^2$ and $A - B = x^2$, we can solve for $A$ and $B$:
   \[
   A = \frac{x^2 + y^2}{2} \quad \text{and} \quad B = \frac{y^2 - x^2}{2}
   \]
6. Since $A$ and $B$ are two-digit numbers, we have:
   \[
   0 \leq A < 100 \quad \text{and} \quad 0 \leq B < 100
   \]
7. Given that $y^2 = kx^2$, we need to find values of $k$ such that $y^2$ and $x^2$ are perfect squares and $A$ and $B$ are two-digit numbers.
8. The possible values of $k$ that satisfy $y^2 = kx^2$ and are perfect squares are $k = 1, 4, 9$. This gives $y = x, 2x, 3x$.
9. For $k = 1$, we have $y = x$:
   \[
   A = \frac{x^2 + x^2}{2} = x^2 \quad \text{and} \quad B = \frac{x^2 - x^2}{2} = 0
   \]
   The possible values of $x$ such that $A$ is a two-digit number are $x = 1, 2, 3, \ldots, 9$. This gives the numbers:
   \[
   0100, 0400, 0900, 1600, 2500, 3600, 4900, 6400, 8100
   \]
10. For $k = 4$, we have $y = 2x$:
    \[
    A = \frac{x^2 + 4x^2}{2} = \frac{5x^2}{2} \quad \text{and} \quad B = \frac{4x^2 - x^2}{2} = \frac{3x^2}{2}
    \]
    The possible values of $x$ such that $A$ and $B$ are two-digit numbers are $x = 2, 4$. This gives the numbers:
    \[
    0504, 2016
    \]
11. For $k = 9$, we have $y = 3x$:
    \[
    A = \frac{x^2 + 9x^2}{2} = 5x^2 \quad \text{and} \quad B = \frac{9x^2 - x^2}{2} = 4x^2
    \]
    The possible values of $x$ such that $A$ and $B$ are two-digit numbers are $x = 1, 2, 3, 4$. This gives the numbers:
    \[
    0504, 2016, 4536, 8064
    \]

The final answer is $\boxed{0100, 0400, 0900, 1600, 2500, 3600, 4900, 6400, 8100, 0504, 2016, 4536, 8064}$