For the natural numbers $$m$$, $$n$$, if $$(1-y)^m(1+y)^n = 1 + a_1 y + a_2 y^2 + \ldots + a_{m+n}y^{m+n}$$ and $$a_1 = a_2 = 10$$, then the value of $$(m+n)$$ is equal to:

We are given $$(1-y)^m(1+y)^n = 1 + a_1 y + a_2 y^2 + \ldots$$ with $$a_1 = a_2 = 10$$.

Expanding to first and second order: $$(1-y)^m = 1 - my + \frac{m(m-1)}{2}y^2 - \ldots$$ $$(1+y)^n = 1 + ny + \frac{n(n-1)}{2}y^2 + \ldots$$

The coefficient of $$y$$ in the product is $$n - m = a_1 = 10$$, so $$n - m = 10$$.

The coefficient of $$y^2$$ is: $$\frac{n(n-1)}{2} - mn + \frac{m(m-1)}{2} = a_2 = 10.$$

Simplifying: $$\frac{n^2 - n - 2mn + m^2 - m}{2} = 10$$, which gives $$\frac{(n-m)^2 - (n+m)}{2} = 10$$ $$(n-m)^2 - (n+m) = 20$$ $$100 - (n+m) = 20$$ $$n + m = 80.$$

Therefore, $$m + n = \boxed{80}$$.