Let for a differentiable function $$f : (0, \infty) \rightarrow \mathbb{R}$$, $$f(x) - f(y) \geq \log_e\left(\frac{x}{y}\right) + x - y, \; \forall x, y \in (0, \infty)$$. Then $$\sum_{n=1}^{20} f'\left(\frac{1}{n^2}\right)$$ is equal to _______.

Given $$f(x) - f(y) \geq \ln\left(\frac{x}{y}\right) + x - y$$ for all $$x, y > 0$$.

Setting $$x = y$$: $$0 \geq 0$$. OK.

Setting $$y = x + h$$ (small $$h > 0$$):

$$f(x) - f(x+h) \geq \ln\left(\frac{x}{x+h}\right) + x - (x+h) = -\ln\left(1+\frac{h}{x}\right) - h$$

Dividing by $$-h$$ (flipping inequality):

$$\frac{f(x+h) - f(x)}{h} \leq \frac{\ln(1+h/x)}{h} + 1$$

As $$h \to 0^+$$: $$f'(x) \leq \frac{1}{x} + 1$$.

Similarly, swapping roles ($$y = x, x = x + h$$):

$$f(x+h) - f(x) \geq \ln\left(\frac{x+h}{x}\right) + h$$

$$\frac{f(x+h)-f(x)}{h} \geq \frac{\ln(1+h/x)}{h} + 1$$

As $$h \to 0^+$$: $$f'(x) \geq \frac{1}{x} + 1$$.

Therefore $$f'(x) = \frac{1}{x} + 1$$.

$$\sum_{n=1}^{20} f'\left(\frac{1}{n^2}\right) = \sum_{n=1}^{20} \left(n^2 + 1\right) = \sum_{n=1}^{20} n^2 + 20 = \frac{20 \times 21 \times 41}{6} + 20 = 2870 + 20 = 2890$$

The answer is $$\boxed{2890}$$.