Let $ m\equal{}\left(abab\right)$ and $ n\equal{}\left(cdcd\right)$ be four-digit numbers in decimal system. If $ m\plus{}n$ is a perfect square, what is the largest value of $ a\cdot b\cdot c\cdot d$?

$\textbf{(A)}\ 392  \qquad\textbf{(B)}\ 420  \qquad\textbf{(C)}\ 588  \qquad\textbf{(D)}\ 600  \qquad\textbf{(E)}\ 750$
1. Let \( m = \overline{abab} \) and \( n = \overline{cdcd} \). These can be expressed as:
   \[
   m = 101 \times \overline{ab} \quad \text{and} \quad n = 101 \times \overline{cd}
   \]
   where \(\overline{ab}\) and \(\overline{cd}\) are two-digit numbers.

2. Given that \( m + n \) is a perfect square, we have:
   \[
   m + n = 101 \times \overline{ab} + 101 \times \overline{cd} = 101 (\overline{ab} + \overline{cd})
   \]
   Let \( \overline{ab} + \overline{cd} = k \). Then:
   \[
   101k = t^2 \quad \text{for some integer } t
   \]
   This implies \( k \) must be a perfect square.

3. Since \(\overline{ab}\) and \(\overline{cd}\) are two-digit numbers, their sum \( k \) must satisfy:
   \[
   20 \leq \overline{ab} + \overline{cd} \leq 198
   \]
   The only perfect square in this range is \( k = 1 \) (since \( 101 \times 1 = 101 \) is the only perfect square that fits).

4. Therefore:
   \[
   \overline{ab} + \overline{cd} = 101
   \]

5. Let \(\overline{ab} = 10a + b\) and \(\overline{cd} = 10c + d\). Then:
   \[
   10a + b + 10c + d = 101
   \]

6. To maximize \( a \cdot b \cdot c \cdot d \), we need to find values of \( a, b, c, d \) such that:
   \[
   10(a + c) + (b + d) = 101
   \]

7. Since \( a, b, c, d \) are digits (1 to 9), we need to find \( a + c \) and \( b + d \) such that:
   \[
   10(a + c) + (b + d) = 101
   \]
   This implies:
   \[
   b + d = 101 - 10(a + c)
   \]

8. To maximize the product \( a \cdot b \cdot c \cdot d \), we need to find the values of \( a, b, c, d \) that are as close to each other as possible. 

9. We find that:
   \[
   a + c = 9 \quad \text{and} \quad b + d = 11
   \]

10. The pairs \((a, c)\) and \((b, d)\) that maximize the product are:
    \[
    \{a, c\} = \{4, 5\} \quad \text{and} \quad \{b, d\} = \{5, 6\}
    \]

11. Therefore, the maximum product is:
    \[
    P_{\text{max}} = 4 \cdot 5 \cdot 5 \cdot 6 = 600
    \]

The final answer is \(\boxed{600}\)