Consider a function f: ℕ → ℝ , satisfying f(1) + 2f(2) + 3f(3) + +...

Given:

$f\left(1\right) + 2f\left(2\right) + 3f\left(3\right) + \ldots + xf\left(x\right) = x\left(x + 1\right)f\left(x\right)$

$⇒ f\left(1\right) + 2f\left(2\right) + 3f\left(3\right) + \ldots + \left(x - 1\right)f\left(x - 1\right) = x\left(x - 1\right)f\left(x - 1\right)$

Now when $x = 2$, then

$f\left(1\right) + 2f\left(2\right) = 6f\left(2\right)$

$⇒ 1 = 4f\left(2\right)$

$⇒ f\left(2\right) = \dfrac{1}{4}$

When $x = 3$,

$f\left(1\right) + 2f\left(2\right) = 9f\left(3\right) ⇒ f\left(3\right) = \dfrac{1}{6}$

When $x = 4$, $f\left(1\right) + 2f\left(2\right) + 3f\left(3\right) = 16f\left(4\right) ⇒ f\left(4\right) = \dfrac{1}{8}$

So,

$f\left(x\right) = \dfrac{1}{2x}$

Hence,

$\dfrac{1}{f\left(2022\right)} + \dfrac{1}{f\left(2028\right)} = 4044 + 4056 = 8100$