Integers $x_1,x_2,\cdots,x_{100}$ satisfy \[ \frac {1}{\sqrt{x_1}} + \frac {1}{\sqrt{x_2}} + \cdots + \frac {1}{\sqrt{x_{100}}} = 20. \]Find $ \displaystyle\prod_{i \ne j} \left( x_i - x_j \right) $.
1. **Assume the integers \( x_1, x_2, \ldots, x_{100} \) are distinct.** 
   - If \( x_i \) are distinct, then \( x_i \) must be at least \( 1, 2, \ldots, 100 \) in some order.
   - The function \( \frac{1}{\sqrt{x}} \) is decreasing, so the sum \( \sum_{i=1}^{100} \frac{1}{\sqrt{x_i}} \) is maximized when \( x_i = i \).

2. **Calculate the upper bound of the sum \( \sum_{i=1}^{100} \frac{1}{\sqrt{i}} \).**
   - We can approximate the sum using an integral:
     \[
     \sum_{i=1}^{100} \frac{1}{\sqrt{i}} < 1 + \int_{1}^{100} \frac{1}{\sqrt{x}} \, dx
     \]
   - Evaluate the integral:
     \[
     \int_{1}^{100} \frac{1}{\sqrt{x}} \, dx = \left[ 2\sqrt{x} \right]_{1}^{100} = 2\sqrt{100} - 2\sqrt{1} = 20 - 2 = 18
     \]
   - Therefore:
     \[
     \sum_{i=1}^{100} \frac{1}{\sqrt{i}} < 1 + 18 = 19
     \]
   - Since \( 19 < 20 \), it is impossible for the sum to reach 20 with all \( x_i \) distinct.

3. **Conclude that there must be at least two equal values among \( x_1, x_2, \ldots, x_{100} \).**
   - If there are at least two equal values, say \( x_i = x_j \) for some \( i \neq j \), then the product \( \prod_{i \ne j} (x_i - x_j) \) will include a factor of \( (x_i - x_j) = 0 \).

4. **Therefore, the product \( \prod_{i \ne j} (x_i - x_j) \) is zero.**

\[
\boxed{0}
\]