Every member, starting from the third one, of two sequences $\{a_n\}$ and $\{b_n\}$ equals to the sum of two preceding ones. First members are: $a_1 = 1, a_2 = 2, b_1 = 2, b_2 = 1$. How many natural numbers are encountered in both sequences (may be on the different places)?
1. We start by defining the sequences $\{a_n\}$ and $\{b_n\}$ based on the given initial conditions and the recurrence relation. Specifically, we have:
   \[
   a_1 = 1, \quad a_2 = 2, \quad a_n = a_{n-1} + a_{n-2} \quad \text{for} \quad n \geq 3
   \]
   \[
   b_1 = 2, \quad b_2 = 1, \quad b_n = b_{n-1} + b_{n-2} \quad \text{for} \quad n \geq 3
   \]

2. We recognize that the sequence $\{a_n\}$ follows the Fibonacci sequence shifted by one index:
   \[
   a_n = F_{n+1}
   \]
   where $F_i$ denotes the $i$-th Fibonacci number.

3. For the sequence $\{b_n\}$, we use the hint provided:
   \[
   b_n = F_n - F_{n-4}
   \]
   This can be verified by induction, but we will use it directly as given.

4. We need to find the natural numbers that appear in both sequences $\{a_n\}$ and $\{b_n\}$. This translates to finding $n$ and $k$ such that:
   \[
   F_{n+1} = F_k - F_{k-4}
   \]

5. We simplify the equation:
   \[
   F_{n+1} = F_k - F_{k-4}
   \]
   Using the properties of Fibonacci numbers, we know:
   \[
   F_k = F_{k-1} + F_{k-2}
   \]
   \[
   F_{k-4} = F_{k-5} + F_{k-6}
   \]
   Therefore:
   \[
   F_k - F_{k-4} = (F_{k-1} + F_{k-2}) - (F_{k-5} + F_{k-6})
   \]

6. We need to check when this equation holds true. For $n > 4$, it is impossible because:
   \[
   F_{n+1} > F_{n-1} + F_{n-3}
   \]
   This inequality holds because the Fibonacci sequence is strictly increasing.

7. Therefore, we only need to check the initial values:
   \[
   F_1 = 1, \quad F_2 = 1, \quad F_3 = 2, \quad F_4 = 3
   \]
   We find that the only common natural numbers in both sequences are $1, 2, 3$.

Conclusion:
\[
\boxed{3}
\]