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 $\boxed{330}$.