Which number is greater:
$$A=\frac{2.00\ldots04}{1.00\ldots04^2+2.00\ldots04},\text{ or }B=\frac{2.00\ldots02}{1.00\ldots02^2+2.00\ldots02},$$where each of the numbers above contains $1998$ zeros?
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 \( \boxed{ B } \)