Let $$c, k \in R$$. If $$f(x) = (c+1)x^2 + (1-c^2)x + 2k$$ and $$f(x+y) = f(x) + f(y) - xy$$, for all $$x, y \in R$$, then the value of $$|2(f(1) + f(2) + f(3) + \ldots + f(20))|$$ is equal to ______.

Given $$f(x) = (c+1)x^2 + (1-c^2)x + 2k$$ and $$f(x+y) = f(x) + f(y) - xy$$ for all $$x, y \in \mathbb{R}$$. Setting $$x = y = 0$$ yields $$f(0) = 2f(0)$$, so $$f(0) = 0$$ and hence $$2k = 0 \Rightarrow k = 0$$.

Thus $$f(x) = (c+1)x^2 + (1-c^2)x$$, and from the functional equation we have $$f(x+y) = (c+1)(x+y)^2 + (1-c^2)(x+y)$$ while $$f(x) + f(y) - xy = (c+1)x^2 + (1-c^2)x + (c+1)y^2 + (1-c^2)y - xy$$.

Comparing the coefficients of $$xy$$ gives $$2(c+1) = -1 \Rightarrow c + 1 = -\frac{1}{2} \Rightarrow c = -\frac{3}{2}$$, and for the linear terms $$1 - c^2 = 1 - \frac{9}{4} = -\frac{5}{4}$$ matches on both sides. Therefore $$f(x) = -\frac{1}{2}x^2 - \frac{5}{4}x$$.

It follows that $$f(n) = -\frac{1}{2}n^2 - \frac{5}{4}n$$ and so $$\sum_{n=1}^{20} f(n) = -\frac{1}{2}\sum_{n=1}^{20} n^2 - \frac{5}{4}\sum_{n=1}^{20} n$$ with $$\sum_{n=1}^{20} n^2 = \frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870$$ and $$\sum_{n=1}^{20} n = \frac{20 \times 21}{2} = 210$$, giving $$\sum_{n=1}^{20} f(n) = -\frac{2870}{2} - \frac{5 \times 210}{4} = -1435 - \frac{1050}{4} = -1435 - 262.5 = -1697.5$$.

The answer is $$\boxed{3395}$$.