We have $x_i >i$ for all $1 \le i \le n$. 
Find the minimum value of $\frac{(\sum_{i=1}^n x_i)^2}{\sum_{i=1}^n \sqrt{x^2_i - i^2}}$
1. We start with the given inequality \( x_i > i \) for all \( 1 \le i \le n \).
2. We need to find the minimum value of the expression \( \frac{(\sum_{i=1}^n x_i)^2}{\sum_{i=1}^n \sqrt{x_i^2 - i^2}} \).

3. Let's denote:
   \[
   S = \sum_{i=1}^n x_i \quad \text{and} \quad T = \sum_{i=1}^n \sqrt{x_i^2 - i^2}
   \]

4. We use the Cauchy-Schwarz inequality in the form:
   \[
   \left( \sum_{i=1}^n a_i b_i \right)^2 \leq \left( \sum_{i=1}^n a_i^2 \right) \left( \sum_{i=1}^n b_i^2 \right)
   \]
   where \( a_i = 1 \) and \( b_i = \sqrt{x_i^2 - i^2} \).

5. Applying the Cauchy-Schwarz inequality:
   \[
   \left( \sum_{i=1}^n \sqrt{x_i^2 - i^2} \right)^2 \leq \left( \sum_{i=1}^n 1^2 \right) \left( \sum_{i=1}^n (x_i^2 - i^2) \right)
   \]
   \[
   T^2 \leq n \sum_{i=1}^n (x_i^2 - i^2)
   \]

6. Next, we use the fact that:
   \[
   S^2 = \left( \sum_{i=1}^n x_i \right)^2 \geq \left( \sum_{i=1}^n \sqrt{x_i^2 - i^2} \right)^2 + \left( \sum_{i=1}^n i \right)^2
   \]
   This follows from the Minkowski inequality.

7. We know:
   \[
   \left( \sum_{i=1}^n i \right)^2 = \left( \frac{n(n+1)}{2} \right)^2
   \]

8. Combining the inequalities, we get:
   \[
   S^2 \geq T^2 + \left( \frac{n(n+1)}{2} \right)^2
   \]

9. Therefore:
   \[
   \frac{S^2}{T} \geq \frac{T^2 + \left( \frac{n(n+1)}{2} \right)^2}{T}
   \]

10. Simplifying the right-hand side:
    \[
    \frac{S^2}{T} \geq T + \frac{\left( \frac{n(n+1)}{2} \right)^2}{T}
    \]

11. To minimize \( \frac{S^2}{T} \), we need to find the equality condition. Equality holds when \( x_i = \sqrt{2}i \) for \( i = 1, 2, \ldots, n \).

12. Substituting \( x_i = \sqrt{2}i \):
    \[
    S = \sum_{i=1}^n \sqrt{2}i = \sqrt{2} \sum_{i=1}^n i = \sqrt{2} \cdot \frac{n(n+1)}{2}
    \]
    \[
    T = \sum_{i=1}^n \sqrt{(\sqrt{2}i)^2 - i^2} = \sum_{i=1}^n \sqrt{2i^2 - i^2} = \sum_{i=1}^n i = \frac{n(n+1)}{2}
    \]

13. Therefore:
    \[
    \frac{S^2}{T} = \frac{\left( \sqrt{2} \cdot \frac{n(n+1)}{2} \right)^2}{\frac{n(n+1)}{2}} = \frac{2 \cdot \left( \frac{n(n+1)}{2} \right)^2}{\frac{n(n+1)}{2}} = n(n+1)
    \]

14. Hence, the minimum value of the given expression is \( n(n+1) \).

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