1. Let $x = 2.00\ldots04$ and $y = 2.00\ldots02$. Note that $x > y$ since $x$ has more trailing zeros before the final digit 4 compared to $y$ which has more trailing zeros before the final digit 2.
2. We need to compare $A = \frac{x}{(x-1)^2 + x}$ and $B = \frac{y}{(y-1)^2 + y}$.
3. Simplify the expressions for $A$ and $B$:
$$A = \frac{x}{x^2 - 2x + 1 + x} = \frac{x}{x^2 - x + 1}$$
$$B = \frac{y}{y^2 - 2y + 1 + y} = \frac{y}{y^2 - y + 1}$$
4. We claim that $\frac{x}{x^2 - x + 1} < \frac{y}{y^2 - y + 1}$.
5. To compare these fractions, we can cross-multiply to obtain:
$$x(y^2 - y + 1) < y(x^2 - x + 1)$$
6. Expanding both sides:
$$xy^2 - xy + x < yx^2 - yx + y$$
7. Rearrange the terms:
$$xy^2 - yx^2 < y - x$$
8. Factor out common terms:
$$xy(y - x) < y - x$$
9. Since $y - x < 0$ (because $x > y$), we can divide both sides by $y - x$ (noting that dividing by a negative number reverses the inequality):
$$xy > 1$$
10. Given that $x$ and $y$ are very large numbers (each containing 1998 zeros), $xy$ is indeed much greater than 1. Therefore, the inequality $xy > 1$ holds true.

Thus, we have shown that:
$$\frac{x}{x^2 - x + 1} < \frac{y}{y^2 - y + 1}$$

The final answer is $\boxed{ B }$