Consider a function $$f : \mathbb{N} \to \mathbb{R}$$, satisfying $$f(1) + 2f(2) + 3f(3) + \ldots + xf(x) = x(x+1)f(x)$$; $$x \geq 2$$ with $$f(1) = 1$$. Then $$\frac{1}{f(2022)} + \frac{1}{f(2028)}$$ is equal to

We are given $$f(1) + 2f(2) + 3f(3) + \ldots + xf(x) = x(x+1)f(x)$$ for $$x \geq 2$$, with $$f(1) = 1$$. Defining $$S(x) = \sum_{k=1}^{x} kf(k)$$ shows that $$S(x) = x(x+1)f(x)$$ for $$x \geq 2$$.

To find $$f(2)$$, observe that $$S(2) = f(1) + 2f(2) = 1 + 2f(2)$$ and $$S(2) = 2 \cdot 3 \cdot f(2) = 6f(2)$$, hence $$1 + 2f(2) = 6f(2) \implies f(2) = \frac{1}{4}$$.

For $$x \geq 3$$, since $$S(x) - S(x-1) = xf(x)$$, with $$S(x) = x(x+1)f(x)$$ and $$S(x-1) = (x-1)xf(x-1)$$, one has $$x(x+1)f(x) - (x-1)xf(x-1) = xf(x)$$. Simplifying gives $$x^2 f(x) = x(x-1)f(x-1)$$ and hence $$xf(x) = (x-1)f(x-1)$$.

Iterating this recurrence yields $$xf(x) = (x-1)f(x-1) = \ldots = 2f(2) = \frac{1}{2}$$, so that $$f(x) = \frac{1}{2x}$$ for all $$x \geq 2$$.

As a verification, $$f(3) = \frac{1}{6}$$, and $$S(3) = 1 + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{6} = 1 + \frac{1}{2} + \frac{1}{2} = 2$$, while $$3 \cdot 4 \cdot f(3) = 12 \cdot \frac{1}{6} = 2$$.

Finally, $$\frac{1}{f(2022)} + \frac{1}{f(2028)} = 2 \times 2022 + 2 \times 2028 = 4044 + 4056 = \boxed{8100}$$.

The correct answer is Option D.