Let S be the set of all integers x satisfying the inequality x^2 -...

First, solve the quadratic inequality to find the set $S$:
$x^2 - 120x + 2000 \le 0$
$(x - 20)(x - 100) \le 0$
Thus, $x \in [20, 100]$. Since $x$ is an integer, $S = \{20, 21, 22, \dots, 100\}$.

The set $A$ consists of elements $x \in S$ such that $xR42$, which means $\text{gcd}(x, 42) > 1$.
Since $42 = 2 \times 3 \times 7$, an element $x \in S$ belongs to $A$ if it is divisible by $2$, $3$, or $7$.

We need to find the sum of all such elements. Let $S_k$ denote the sum of multiples of $k$ in the set $S$. By the Principle of Inclusion-Exclusion:
$\text{Sum}(A) = S_2 + S_3 + S_7 - S_6 - S_{14} - S_{21} + S_{42}$

Sum of multiples of $2$ in $S$ ($20$ to $100$):
Number of terms $= \dfrac{100 - 20}{2} + 1 = 41$
$S_2 = \dfrac{41}{2}(20 + 100) = 2460$

Sum of multiples of $3$ in $S$ ($21$ to $99$):
Number of terms $= \dfrac{99 - 21}{3} + 1 = 27$
$S_3 = \dfrac{27}{2}(21 + 99) = 1620$

Sum of multiples of $7$ in $S$ ($21$ to $98$):
Number of terms $= \dfrac{98 - 21}{7} + 1 = 12$
$S_7 = \dfrac{12}{2}(21 + 98) = 714$

Sum of multiples of $6$ in $S$ ($24$ to $96$):
Number of terms $= \dfrac{96 - 24}{6} + 1 = 13$
$S_6 = \dfrac{13}{2}(24 + 96) = 780$

Sum of multiples of $14$ in $S$ ($28$ to $98$):
Number of terms $= \dfrac{98 - 28}{14} + 1 = 6$
$S_{14} = \dfrac{6}{2}(28 + 98) = 378$

Sum of multiples of $21$ in $S$ ($21$ to $84$):
Number of terms $= \dfrac{84 - 21}{21} + 1 = 4$
$S_{21} = \dfrac{4}{2}(21 + 84) = 210$

Sum of multiples of $42$ in $S$ ($42$ to $84$):
Number of terms $= \dfrac{84 - 42}{42} + 1 = 2$
$S_{42} = \dfrac{2}{2}(42 + 84) = 126$

Now, substituting these into the inclusion-exclusion formula:
$\text{Sum}(A) = 2460 + 1620 + 714 - 780 - 378 - 210 + 126$
$\text{Sum}(A) = 4794 - 1368 + 126 = 3552$

Answer: $3552$