The numbers $1,2,\cdots,2021$ are arranged in a circle. For any $1 \le i \le 2021$, if $i,i+1,i+2$ are three consecutive numbers in some order such that $i+1$ is not in the middle, then $i$ is said to be a good number. Indices are taken mod $2021$. What is the maximum possible number of good numbers?

[i]CSJL[/i]
1. **Understanding the Problem:**
   We need to arrange the numbers \(1, 2, \ldots, 2021\) in a circle such that for any \(1 \leq i \leq 2021\), if \(i, i+1, i+2\) are three consecutive numbers in some order, \(i+1\) is not in the middle. We need to find the maximum number of such "good" numbers.

2. **Initial Claim:**
   We claim that the maximum number of good numbers is \(2 \times 673 = 1346\). To support this, we provide a construction: \(2020, 1, 2021, 2, 4, 3, 5, 7, 6, \ldots, 2018, 2019\).

3. **Definition of Good Set:**
   A set of three consecutive numbers in the circle containing \(i\) and not having the middle number in the middle is called an \(i\)-good set.

4. **Analysis of Good Sets:**
   - For \(1 \leq i \leq 2019\), the set \(\{i, i+1, i+2\}\) contains a number which is \(0 \pmod{3}\).
   - Given \(i\), there are three sets of consecutive numbers containing \(i\). Two of these sets contain \(i-1\) or \(i+1\). If there were three \(i\)-good sets, then \(i-1\) and \(i+1\) would have to be directly to the left and right of \(i\), making \(\{i-1, i, i+1\}\) not \(i\)-good.
   - Therefore, there are at most two \(i\)-good sets for each \(i\).

5. **Counting Good Numbers:**
   - If there are no good sets not containing a multiple of \(3\), then we are done. There are \(673\) multiples of \(3\) less than or equal to \(2021\), implying that there are at most \(673 \times 2 = 1346\) good numbers.

6. **Possible Good Sets Not Containing a Multiple of 3:**
   - These sets contain two numbers either \(1 \pmod{3}\) or \(2 \pmod{3}\). Without loss of generality, assume one of them is \(2 \pmod{3}\).
   - The possible sets are \(\{1, 2021, 2\}\) and \(\{2, 2021, 1\}\). Every good set contains a number which is \(2 \pmod{3}\), so since both of these sets contain one twice, every other good set has to contain one exactly once.

7. **Constructing the Sequence:**
   - Assume the set is \(\{2, 2021, 1\}\). The only possible good set for \(2\) is \(\{3, 4, 2\}\).
   - Then \(5\) must be part of two good sets, one of which must contain \(4\). The only way to place \(5\) is \(\{5, 3, 4, 2, 2021, 1\}\).
   - Continue this pattern for \(6, 7, \ldots, 2018\). Eventually, we cannot place numbers like \(\{2018, 2020, 2019\}\) because \(1\) is in front of \(2020\).

8. **Conclusion:**
   - The maximum number of good numbers is \(1346\).

The final answer is \(\boxed{1346}\)