The integer sequence $(s_i)$ "having pattern 2016'" is defined as follows:\\
$\circ$ The first member $s_1$ is 2.\\
$\circ$ The second member $s_2$ is the least positive integer exceeding $s_1$ and having digit 0 in its decimal notation.\\
$\circ$ The third member $s_3$ is the least positive integer exceeding $s_2$ and having digit 1 in its decimal notation.\\
$\circ$ The third member $s_3$ is the least positive integer exceeding $s_2$ and having digit 6 in its decimal notation.\\
The following members are defined in the same way. The required digits change periodically: $2 \rightarrow 0 \rightarrow 1 \rightarrow 6 \rightarrow 2 \rightarrow 0 \rightarrow \ldots$. The first members of this sequence are the following: $2; 10; 11; 16; 20; 30; 31; 36; 42; 50$. What are the 4 numbers that immediately follow $s_k = 2016$ in this sequence?
To solve this problem, we need to follow the pattern defined in the sequence and find the four numbers that immediately follow \( s_k = 2016 \).

1. **Identify the pattern and the position of \( 2016 \) in the sequence:**
   The sequence follows the pattern \( 2 \rightarrow 0 \rightarrow 1 \rightarrow 6 \rightarrow 2 \rightarrow 0 \rightarrow \ldots \). We need to determine the position \( k \) of \( 2016 \) in the sequence. 

   Since \( 2016 \) ends with the digit 6, it must be at a position where the required digit is 6. The pattern repeats every 4 steps, so \( k \equiv 0 \pmod{4} \).

2. **Determine the next four numbers in the sequence:**
   Given \( s_k = 2016 \) and \( k \equiv 0 \pmod{4} \), the next required digits will follow the pattern:
   - \( s_{k+1} \) must have the digit 2.
   - \( s_{k+2} \) must have the digit 0.
   - \( s_{k+3} \) must have the digit 1.
   - \( s_{k+4} \) must have the digit 6.

3. **Find the next four numbers:**
   - \( s_{k+1} \): The smallest number greater than 2016 that contains the digit 2 is \( 2017 \).
   - \( s_{k+2} \): The smallest number greater than 2017 that contains the digit 0 is \( 2020 \).
   - \( s_{k+3} \): The smallest number greater than 2020 that contains the digit 1 is \( 2021 \).
   - \( s_{k+4} \): The smallest number greater than 2021 that contains the digit 6 is \( 2026 \).

Therefore, the four numbers that immediately follow \( s_k = 2016 \) in the sequence are \( 2017, 2020, 2021, \) and \( 2026 \).

The final answer is \(\boxed{2017, 2020, 2021, 2026}\).