Find all pairs of coprime natural numbers $a$ and $b$ such that the fraction $\frac{a}{b}$ is written in the decimal system as $b.a.$
To solve the problem, we need to find all pairs of coprime natural numbers \(a\) and \(b\) such that the fraction \(\frac{a}{b}\) is written in the decimal system as \(b.a\). This means that:

\[
\frac{a}{b} = b + \frac{a}{10^n}
\]

where \(10^{n-1} \leq a < 10^n\) and \(1 \leq b\).

1. **Rewrite the equation:**

   \[
   \frac{a}{b} = b + \frac{a}{10^n}
   \]

   Multiply both sides by \(10^n\):

   \[
   10^n \cdot \frac{a}{b} = 10^n b + a
   \]

   Simplify:

   \[
   10^n a = b (10^n b + a)
   \]

2. **Rearrange the equation:**

   \[
   10^n a = 10^n b^2 + ab
   \]

   \[
   10^n a - ab = 10^n b^2
   \]

   Factor out \(a\) on the left-hand side:

   \[
   a (10^n - b) = 10^n b^2
   \]

3. **Analyze the divisibility:**

   Since \(a\) and \(b\) are coprime, \(a\) must divide \(10^n\). Similarly, \(b\) must divide \(10^n\). Therefore, \(a\) and \(b\) must be powers of 2 and 5, respectively, or vice versa.

4. **Case 1: \(a = 2^u\) and \(b = 5^v\)**

   Substitute \(a = 2^u\) and \(b = 5^v\) into the equation:

   \[
   2^u (10^n - 5^v) = 10^n 5^{2v}
   \]

   Simplify:

   \[
   2^u 10^n - 2^u 5^v = 10^n 5^{2v}
   \]

   \[
   2^u 10^n = 10^n 5^{2v} + 2^u 5^v
   \]

   Since \(10^n = 2^n \cdot 5^n\), we can rewrite the equation as:

   \[
   2^u 2^n 5^n = 2^n 5^{n+2v} + 2^u 5^v
   \]

   \[
   2^{u+n} 5^n = 2^n 5^{n+2v} + 2^u 5^v
   \]

   For the equation to hold, we must have \(u = n\) and \(v = n\). However, this leads to no solution as shown in the initial solution.

5. **Case 2: \(a = 5^u\) and \(b = 2^v\)**

   Substitute \(a = 5^u\) and \(b = 2^v\) into the equation:

   \[
   5^u (10^n - 2^v) = 10^n 2^{2v}
   \]

   Simplify:

   \[
   5^u 10^n - 5^u 2^v = 10^n 2^{2v}
   \]

   \[
   5^u 10^n = 10^n 2^{2v} + 5^u 2^v
   \]

   Since \(10^n = 2^n \cdot 5^n\), we can rewrite the equation as:

   \[
   5^u 2^n 5^n = 2^n 5^n 2^{2v} + 5^u 2^v
   \]

   \[
   5^{u+n} 2^n = 2^n 5^n 2^{2v} + 5^u 2^v
   \]

   For the equation to hold, we must have \(u = n\) and \(v = n\). This leads to the solution:

   \[
   a = 5, \quad b = 2
   \]

   Verify the solution:

   \[
   \frac{5}{2} = 2.5
   \]

   This matches the condition given in the problem.

\(\blacksquare\)

The final answer is \( \boxed{ (a, b) = (5, 2) } \)