Let k be a natural number such that 1 ≤ k ≤ 20 . Consider a...

For $R$ to be an equivalence relation, it must be reflexive.

For $R$ to be reflexive, $xRx$ must hold for all $x \in \mathbb{N}$.
$\Rightarrow (k^2 - 5)x + (2k + 1)x \text{ is a multiple of }11$ for all $x \in \mathbb{N}$.
$\Rightarrow (k^2 + 2k - 4)x \text{ is a multiple of }11$ for all $x \in \mathbb{N}$.

This requires $k^2 + 2k - 4$ to be a multiple of $11$.
$\Rightarrow k^2 + 2k - 4 \equiv 0 \pmod{11}$
Adding $5$ to both sides to complete the square:
$\Rightarrow k^2 + 2k + 1 \equiv 5 \pmod{11}$
$\Rightarrow (k + 1)^2 \equiv 5 \pmod{11}$

Checking the squares modulo $11$:
$1^2 = 1$
$2^2 = 4$
$3^2 = 9$
$4^2 = 16 \equiv 5 \pmod{11}$
$5^2 = 25 \equiv 3 \pmod{11}$

Thus, $k + 1 \equiv 4 \pmod{11}$ or $k + 1 \equiv -4 \equiv 7 \pmod{11}$.
This gives $k \equiv 3 \pmod{11}$ or $k \equiv 6 \pmod{11}$.

For these values, $(2k + 1) \equiv -(k^2 - 5) \pmod{11}$, and the relation reduces to $(k^2 - 5)(x - y) \equiv 0 \pmod{11}$, which is symmetric and transitive.

The possible values of $k$ in the interval $[1, 20]$ are:
For $k \equiv 3 \pmod{11}$: $k = 3, 14$
For $k \equiv 6 \pmod{11}$: $k = 6, 17$

The sum of all possible values of $k$ is $3 + 14 + 6 + 17 = 40$.

Answer: $40$