Let $$S$$ be the set of all the natural numbers, for which the line $$\frac{x}{a} + \frac{y}{b} = 2$$ is a tangent to the curve $$\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$$ at the point $$(a, b), ab \neq 0$$. Then

We need to find the set $$S$$ of natural numbers $$n$$ for which the line $$\frac{x}{a} + \frac{y}{b} = 2$$ is tangent to the curve $$\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$$ at the point $$(a, b)$$.

At $$(a, b)$$ the curve gives $$1^n + 1^n = 2$$ $$\checkmark$$ and the line gives $$1 + 1 = 2$$ $$\checkmark$$.

Next, differentiate $$\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$$ implicitly to get
$$\frac{n}{a}\left(\frac{x}{a}\right)^{n-1} + \frac{n}{b}\left(\frac{y}{b}\right)^{n-1}\frac{dy}{dx} = 0$$ Substituting $$(x,y)=(a,b)$$ yields $$\frac{n}{a}(1) + \frac{n}{b}(1)\frac{dy}{dx} = 0$$, so that $$\frac{dy}{dx} = -\frac{b}{a}$$.

Then the tangent line at $$(a,b)$$ satisfies $$y - b = -\frac{b}{a}(x - a)$$. It follows that $$y = -\frac{b}{a}x + b + b = -\frac{b}{a}x + 2b$$, which leads to $$\frac{b}{a}x + y = 2b$$ and hence $$\frac{x}{a} + \frac{y}{b} = 2$$.

Thus the tangent at $$(a, b)$$ to the curve is $$\frac{x}{a} + \frac{y}{b} = 2$$ for every natural number $$n$$, since the slope $$-\frac{b}{a}$$ is independent of $$n$$. Therefore, the given line is tangent to the curve at $$(a, b)$$ for all $$n \in \mathbb{N}$$.

The correct answer is Option D: $$S = \mathbb{N}$$.