Let $a$, $b$, $c$, and $d$ be positive real numbers such that
\[a^2 + b^2 - c^2 - d^2 = 0 \quad \text{and} \quad a^2 - b^2 - c^2 + d^2 = \frac{56}{53}(bc + ad).\]
Let $M$ be the maximum possible value of $\tfrac{ab+cd}{bc+ad}$. If $M$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, find $100m + n$.

[i]Proposed by Robin Park[/i]
1. **Given Equations:**
   We start with the given equations:
   \[
   a^2 + b^2 - c^2 - d^2 = 0
   \]
   \[
   a^2 - b^2 - c^2 + d^2 = \frac{56}{53}(bc + ad)
   \]

2. **Trigonometric Substitution:**
   To simplify these equations, we use trigonometric identities. Assume:
   \[
   a = r \sin x, \quad b = r \cos x, \quad c = r \sin y, \quad d = r \cos y
   \]
   where \( r \) is a positive real number.

3. **Substituting into the First Equation:**
   Substitute \( a, b, c, \) and \( d \) into the first equation:
   \[
   (r \sin x)^2 + (r \cos x)^2 - (r \sin y)^2 - (r \cos y)^2 = 0
   \]
   Simplifying, we get:
   \[
   r^2 (\sin^2 x + \cos^2 x - \sin^2 y - \cos^2 y) = 0
   \]
   Since \(\sin^2 \theta + \cos^2 \theta = 1\) for any angle \(\theta\), this reduces to:
   \[
   r^2 (1 - 1) = 0
   \]
   This equation is always true, so it does not provide additional constraints.

4. **Substituting into the Second Equation:**
   Substitute \( a, b, c, \) and \( d \) into the second equation:
   \[
   (r \sin x)^2 - (r \cos x)^2 - (r \sin y)^2 + (r \cos y)^2 = \frac{56}{53} (r \cos x \cdot r \sin y + r \sin x \cdot r \cos y)
   \]
   Simplifying, we get:
   \[
   r^2 (\sin^2 x - \cos^2 x - \sin^2 y + \cos^2 y) = \frac{56}{53} r^2 (\cos x \sin y + \sin x \cos y)
   \]
   Dividing both sides by \( r^2 \):
   \[
   \sin^2 x - \cos^2 x - \sin^2 y + \cos^2 y = \frac{56}{53} (\cos x \sin y + \sin x \cos y)
   \]

5. **Using Trigonometric Identities:**
   Recognize that:
   \[
   \sin^2 x - \cos^2 x = -\cos 2x \quad \text{and} \quad \sin^2 y - \cos^2 y = -\cos 2y
   \]
   Thus, the equation becomes:
   \[
   -\cos 2x + \cos 2y = \frac{56}{53} \sin(x + y)
   \]

6. **Simplifying Further:**
   Using the identity for the difference of cosines:
   \[
   \cos 2y - \cos 2x = 2 \sin(x + y) \sin(x - y)
   \]
   We get:
   \[
   2 \sin(x + y) \sin(x - y) = \frac{56}{53} \sin(x + y)
   \]
   Assuming \(\sin(x + y) \neq 0\), we can divide both sides by \(\sin(x + y)\):
   \[
   2 \sin(x - y) = \frac{56}{53}
   \]
   Thus:
   \[
   \sin(x - y) = \frac{28}{53}
   \]

7. **Finding \(\cos(x - y)\):**
   Using the Pythagorean identity:
   \[
   \cos^2(x - y) + \sin^2(x - y) = 1
   \]
   We have:
   \[
   \cos^2(x - y) = 1 - \left(\frac{28}{53}\right)^2 = 1 - \frac{784}{2809} = \frac{2025}{2809}
   \]
   Therefore:
   \[
   \cos(x - y) = \frac{45}{53}
   \]

8. **Maximizing the Expression:**
   We need to maximize:
   \[
   \frac{ab + cd}{bc + ad}
   \]
   Substituting \( a, b, c, \) and \( d \):
   \[
   \frac{r \sin x \cdot r \cos x + r \sin y \cdot r \cos y}{r \cos x \cdot r \sin y + r \sin x \cdot r \cos y} = \frac{\sin x \cos x + \sin y \cos y}{\cos x \sin y + \sin x \cos y}
   \]
   Using the double-angle identities:
   \[
   \frac{\frac{1}{2} \sin 2x + \frac{1}{2} \sin 2y}{\sin(x + y)} = \frac{\sin 2x + \sin 2y}{2 \sin(x + y)}
   \]
   Using the sum-to-product identities:
   \[
   \frac{2 \sin(x + y) \cos(x - y)}{2 \sin(x + y)} = \cos(x - y)
   \]
   Thus, the maximum value is:
   \[
   \cos(x - y) = \frac{45}{53}
   \]

The final answer is \( \boxed{ 100m + n = 100 \times 45 + 53 = 4553 } \).