Let $$K$$ be the sum of the coefficients of the odd powers of $$x$$ in the expansion of $$(1+x)^{99}$$. Let $$a$$ be the middle term in the expansion of $$\left(2 + \frac{1}{\sqrt{2}}\right)^{200}$$. If $$\frac{^{200}C_{99}K}{a} = \frac{2^l m}{n}$$, where $$m$$ and $$n$$ are odd numbers, then the ordered pair $$(l, n)$$ is equal to:

We need to find the ordered pair $$(l, n)$$ given the expression $$\dfrac{{}^{200}C_{99} \cdot K}{a} = \dfrac{2^l \cdot m}{n}$$ where $$m$$ and $$n$$ are odd.

Put $$x = 1$$: $$(1+1)^{99} = 2^{99}$$ (sum of all coefficients).

Put $$x = -1$$: $$(1-1)^{99} = 0$$ (alternating sum).

Sum of coefficients of odd powers = $$\dfrac{2^{99} - 0}{2} = 2^{98}$$.

So $$K = 2^{98}$$.

The expansion has 201 terms, so the middle term is the 101st term ($$r = 100$$):

$$a = {}^{200}C_{100} \cdot 2^{100} \cdot \left(\frac{1}{\sqrt{2}}\right)^{100} = {}^{200}C_{100} \cdot 2^{100} \cdot 2^{-50} = {}^{200}C_{100} \cdot 2^{50}$$

$$\frac{{}^{200}C_{99} \cdot K}{a} = \frac{{}^{200}C_{99} \cdot 2^{98}}{{}^{200}C_{100} \cdot 2^{50}}$$

Now, $$\dfrac{{}^{200}C_{99}}{{}^{200}C_{100}} = \dfrac{100!\ \cdot\ 100!}{99!\ \cdot\ 101!} = \dfrac{100}{101}$$.

$$= \frac{100}{101} \cdot 2^{48}$$

$$100 = 2^2 \times 25$$, so:

$$\frac{100}{101} \cdot 2^{48} = \frac{25 \cdot 2^2 \cdot 2^{48}}{101} = \frac{2^{50} \cdot 25}{101}$$

Here $$m = 25$$ (odd) and $$n = 101$$ (odd), and $$l = 50$$.

The ordered pair $$(l, n) = (50, 101)$$.

The correct answer is Option C: $$(50, 101)$$.