Find the sum of the real roots of the polynomial \[ \prod_{k=1}^{100} \left( x^2-11x+k \right) = \left( x^2-11x+1 \right)\left( x^2-11x+2 \right)\dots\left(x^2-11x+100\right). \][i]Proposed by Evan Chen[/i]
To find the sum of the real roots of the polynomial 
\[ \prod_{k=1}^{100} \left( x^2-11x+k \right) = \left( x^2-11x+1 \right)\left( x^2-11x+2 \right)\dots\left(x^2-11x+100\right), \]
we need to analyze the quadratic factors \( x^2 - 11x + k \) for \( k = 1, 2, \ldots, 100 \).

1. **Determine the conditions for real roots:**
   Each quadratic \( x^2 - 11x + k \) has real roots if and only if its discriminant is non-negative. The discriminant of \( x^2 - 11x + k \) is given by:
   \[
   \Delta = b^2 - 4ac = 11^2 - 4 \cdot 1 \cdot k = 121 - 4k.
   \]
   For the quadratic to have real roots, we need:
   \[
   121 - 4k \geq 0 \implies 121 \geq 4k \implies k \leq \frac{121}{4} = 30.25.
   \]
   Since \( k \) must be an integer, the largest possible value of \( k \) is 30. Therefore, the quadratics \( x^2 - 11x + k \) have real roots for \( k = 1, 2, \ldots, 30 \).

2. **Sum of the roots of each quadratic:**
   By Vieta's formulas, the sum of the roots of the quadratic equation \( x^2 - 11x + k = 0 \) is given by:
   \[
   \alpha + \beta = 11.
   \]
   This is true for each \( k \) from 1 to 30.

3. **Total sum of the real roots:**
   Since there are 30 such quadratics, and each quadratic contributes a sum of 11 for its roots, the total sum of the real roots is:
   \[
   30 \times 11 = 330.
   \]

The final answer is \(\boxed{330}\).