Let $a$ be a real number greater than $1$ such that $\frac{20a}{a^2+1} = \sqrt{2}$. Find $\frac{14a}{a^2 - 1}$.
1. Given the equation:
   \[
   \frac{20a}{a^2 + 1} = \sqrt{2}
   \]
   We need to find the value of \(\frac{14a}{a^2 - 1}\).

2. First, solve for \(a\) from the given equation. Multiply both sides by \(a^2 + 1\):
   \[
   20a = \sqrt{2}(a^2 + 1)
   \]
   Rearrange the equation:
   \[
   20a = \sqrt{2}a^2 + \sqrt{2}
   \]
   \[
   \sqrt{2}a^2 - 20a + \sqrt{2} = 0
   \]
   This is a quadratic equation in terms of \(a\). Let \(x = a\):
   \[
   \sqrt{2}x^2 - 20x + \sqrt{2} = 0
   \]

3. Solve the quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
   \[
   x = \frac{20 \pm \sqrt{400 - 4 \cdot \sqrt{2} \cdot \sqrt{2}}}{2 \cdot \sqrt{2}}
   \]
   \[
   x = \frac{20 \pm \sqrt{400 - 8}}{2 \cdot \sqrt{2}}
   \]
   \[
   x = \frac{20 \pm \sqrt{392}}{2 \cdot \sqrt{2}}
   \]
   \[
   x = \frac{20 \pm 14\sqrt{2}}{2 \cdot \sqrt{2}}
   \]
   \[
   x = \frac{20}{2\sqrt{2}} \pm \frac{14\sqrt{2}}{2\sqrt{2}}
   \]
   \[
   x = \frac{10}{\sqrt{2}} \pm 7
   \]
   \[
   x = 5\sqrt{2} \pm 7
   \]

4. Since \(a > 1\), we choose the positive root:
   \[
   a = 5\sqrt{2} + 7
   \]

5. Now, we need to find \(\frac{14a}{a^2 - 1}\). First, calculate \(a^2\):
   \[
   a = 5\sqrt{2} + 7
   \]
   \[
   a^2 = (5\sqrt{2} + 7)^2
   \]
   \[
   a^2 = (5\sqrt{2})^2 + 2 \cdot 5\sqrt{2} \cdot 7 + 7^2
   \]
   \[
   a^2 = 50 + 70\sqrt{2} + 49
   \]
   \[
   a^2 = 99 + 70\sqrt{2}
   \]

6. Calculate \(a^2 - 1\):
   \[
   a^2 - 1 = 99 + 70\sqrt{2} - 1
   \]
   \[
   a^2 - 1 = 98 + 70\sqrt{2}
   \]

7. Now, calculate \(\frac{14a}{a^2 - 1}\):
   \[
   \frac{14a}{a^2 - 1} = \frac{14(5\sqrt{2} + 7)}{98 + 70\sqrt{2}}
   \]
   Simplify the numerator:
   \[
   14(5\sqrt{2} + 7) = 70\sqrt{2} + 98
   \]
   Thus:
   \[
   \frac{14a}{a^2 - 1} = \frac{70\sqrt{2} + 98}{98 + 70\sqrt{2}}
   \]
   Notice that the numerator and the denominator are the same:
   \[
   \frac{70\sqrt{2} + 98}{70\sqrt{2} + 98} = 1
   \]

The final answer is \(\boxed{1}\).