Let's find a positive real number that becomes 2501 times larger when the first and fifth digits after the decimal point in its decimal form are swapped.
Solution. Let the sought number be $x=\overline{A, b c d e f g h i \ldots}$, and the number obtained after the swap be $y=\overline{A, f c d e b g h i \ldots}$ (where $A$ is the integer part of the number). Since $\frac{y'}{x}<2$ when $A>0$, and $y$ is 2501 times $x$, the value of $A$ must be 0. Therefore, the value of $y$ is less than 1, which means $x<\frac{1}{2501}=0.000399984 \ldots$. Thus, the values of $b, c, d$ are also 0.

Writing the difference $y-x=\overline{A, f c d e b g h i \ldots}-\overline{A, b c d e f g h i \ldots}$ and using the fact that $b=0$:

$$
y-x=2500 x=\frac{f}{10}-\frac{f}{100000},
$$

from which

$$
x=\frac{\frac{9999}{100000}}{2500} f=\frac{9999}{250000000} f.
$$

By checking the possible values of $f$ (0, 1, 2, ..., 9), we find that only $f=7$ fits. Therefore, the value of the sought real number is:

$$
x=\frac{9999}{250000000} \cdot 7=0.000279972
$$

(The value of $y$ is 0.700209972).