Find a positive integrer number $n$ such that, if yor put a number $2$ on the left and a number $1$ on the right, the new number is equal to $33n$.
1. Let \( n \) be a positive integer with \( k \) digits. We need to find \( n \) such that placing a 2 on the left and a 1 on the right of \( n \) results in a number equal to \( 33n \).

2. The number formed by placing 2 on the left and 1 on the right of \( n \) can be expressed as \( 2 \cdot 10^{k+1} + 10n + 1 \).

3. According to the problem, this new number should be equal to \( 33n \). Therefore, we have the equation:
   \[
   2 \cdot 10^{k+1} + 10n + 1 = 33n
   \]

4. Rearrange the equation to isolate \( n \):
   \[
   2 \cdot 10^{k+1} + 1 = 33n - 10n
   \]
   \[
   2 \cdot 10^{k+1} + 1 = 23n
   \]
   \[
   n = \frac{2 \cdot 10^{k+1} + 1}{23}
   \]

5. To find \( k \), we need \( 2 \cdot 10^{k+1} + 1 \) to be divisible by 23. This implies:
   \[
   2 \cdot 10^{k+1} \equiv -1 \pmod{23}
   \]
   \[
   10^{k+1} \equiv 11 \pmod{23}
   \]

6. We need to find the smallest \( k \) such that \( 10^{k+1} \equiv 11 \pmod{23} \). We can test values of \( k \):

   - For \( k = 1 \):
     \[
     10^2 \equiv 100 \equiv 8 \pmod{23} \quad (\text{not } 11)
     \]

   - For \( k = 2 \):
     \[
     10^3 \equiv 1000 \equiv 11 \pmod{23} \quad (\text{this works})
     \]

7. Since \( k = 2 \) works, we substitute \( k = 2 \) back into the equation for \( n \):
   \[
   n = \frac{2 \cdot 10^{2+1} + 1}{23}
   \]
   \[
   n = \frac{2 \cdot 10^3 + 1}{23}
   \]
   \[
   n = \frac{2000 + 1}{23}
   \]
   \[
   n = \frac{2001}{23}
   \]
   \[
   n = 87
   \]

Conclusion:
\[
\boxed{87}
\]