Find all $ 3$-digit numbers such that placing to the right side of the number its successor we get a $ 6$-digit number which is a perfect square.
1. Let \( n \) be a 3-digit number. We need to find \( n \) such that placing its successor \( n+1 \) to the right of \( n \) forms a 6-digit number which is a perfect square. This can be written as:
   \[
   1000n + (n+1) = 1001n + 1
   \]
   We need \( 1001n + 1 \) to be a perfect square. Let \( m^2 = 1001n + 1 \), where \( m \) is an integer.

2. Rearrange the equation to solve for \( n \):
   \[
   n = \frac{m^2 - 1}{1001}
   \]
   For \( n \) to be an integer, \( m^2 - 1 \) must be divisible by 1001. Note that \( 1001 = 7 \times 11 \times 13 \). Therefore, \( m^2 \equiv 1 \pmod{1001} \).

3. We need to find \( m \) such that:
   \[
   m^2 \equiv 1 \pmod{7}, \quad m^2 \equiv 1 \pmod{11}, \quad m^2 \equiv 1 \pmod{13}
   \]
   This implies:
   \[
   m \equiv \pm 1 \pmod{7}, \quad m \equiv \pm 1 \pmod{11}, \quad m \equiv \pm 1 \pmod{13}
   \]

4. We now consider all combinations of \( m \) modulo 7, 11, and 13:
   - \( m \equiv 1 \pmod{7} \)
   - \( m \equiv -1 \pmod{7} \)
   - \( m \equiv 1 \pmod{11} \)
   - \( m \equiv -1 \pmod{11} \)
   - \( m \equiv 1 \pmod{13} \)
   - \( m \equiv -1 \pmod{13} \)

5. Using the Chinese Remainder Theorem, we solve for \( m \) in each case:
   - \( m \equiv 1 \pmod{7} \), \( m \equiv 1 \pmod{11} \), \( m \equiv 1 \pmod{13} \):
     \[
     m \equiv 1 \pmod{1001} \quad \text{(not in } [317, 999])
     \]
   - \( m \equiv 1 \pmod{7} \), \( m \equiv 1 \pmod{11} \), \( m \equiv -1 \pmod{13} \):
     \[
     m \equiv 155 \pmod{1001} \quad \text{(not in } [317, 999])
     \]
   - \( m \equiv 1 \pmod{7} \), \( m \equiv -1 \pmod{11} \), \( m \equiv 1 \pmod{13} \):
     \[
     m \equiv 274 \pmod{1001} \quad \text{(not in } [317, 999])
     \]
   - \( m \equiv 1 \pmod{7} \), \( m \equiv -1 \pmod{11} \), \( m \equiv -1 \pmod{13} \):
     \[
     m \equiv 428 \pmod{1001} \quad \text{(in } [317, 999])
     \]
     \[
     n = \frac{428^2 - 1}{1001} = 183
     \]
   - \( m \equiv -1 \pmod{7} \), \( m \equiv 1 \pmod{11} \), \( m \equiv 1 \pmod{13} \):
     \[
     m \equiv 573 \pmod{1001} \quad \text{(in } [317, 999])
     \]
     \[
     n = \frac{573^2 - 1}{1001} = 328
     \]
   - \( m \equiv -1 \pmod{7} \), \( m \equiv 1 \pmod{11} \), \( m \equiv -1 \pmod{13} \):
     \[
     m \equiv 727 \pmod{1001} \quad \text{(in } [317, 999])
     \]
     \[
     n = \frac{727^2 - 1}{1001} = 528
     \]
   - \( m \equiv -1 \pmod{7} \), \( m \equiv -1 \pmod{11} \), \( m \equiv 1 \pmod{13} \):
     \[
     m \equiv 846 \pmod{1001} \quad \text{(in } [317, 999])
     \]
     \[
     n = \frac{846^2 - 1}{1001} = 715
     \]
   - \( m \equiv -1 \pmod{7} \), \( m \equiv -1 \pmod{11} \), \( m \equiv -1 \pmod{13} \):
     \[
     m \equiv 1000 \pmod{1001} \quad \text{(not in } [317, 999])
     \]

6. Therefore, the valid 3-digit numbers \( n \) are:
   \[
   n \in \{183, 328, 528, 715\}
   \]

The final answer is \(\boxed{n \in \{183, 328, 528, 715\}}\)