1.	Around a circumference are written $2012$ number, each of with is equal to $1$ or $-1$. If there are not $10$ consecutive numbers that sum $0$, find all possible values of the sum of the $2012$ numbers.
1. **Label the numbers**: Let the numbers be \( a_1, a_2, \dots, a_{2012} \), where each \( a_i \) is either \( 1 \) or \( -1 \).

2. **Define the sum of 10 consecutive numbers**: Let \( S(i) \) be the sum of the ten consecutive numbers starting from \( a_i \), i.e., 
   \[
   S(i) = a_i + a_{i+1} + \cdots + a_{i+9}
   \]
   Note that \( S(i) \) is even because it is the sum of an even number of \( \pm 1 \) values.

3. **Properties of \( S(i) \)**: Since \( S(i) \) is even, the difference between consecutive sums \( S(i) \) and \( S(i+1) \) must be \( 0 \) or \( \pm 2 \):
   \[
   |S(i) - S(i+1)| \in \{0, 2\}
   \]

4. **Intermediate Value Theorem**: If there existed \( i \) and \( j \) such that \( S(i) \) and \( S(j) \) had opposite signs, then by the intermediate value theorem, there would be some \( k \) such that \( S(k) = 0 \). However, we are given that no 10 consecutive numbers sum to 0, so \( S(i) \) must be either all positive or all negative.

5. **Assume \( S(i) \) are all positive**: Without loss of generality, assume \( S(i) \) are all positive. The sum of all \( S(i) \) over all \( i \) must be at least \( 2 \times 2012 = 4024 \) because each \( S(i) \) is at least 2.

6. **Counting each number 10 times**: Since each number \( a_i \) is counted in exactly 10 different \( S(i) \), the total sum of all \( a_i \) must be at least:
   \[
   \left\lceil \frac{4024}{10} \right\rceil = 403
   \]
   However, since the sum of all \( a_i \) must be even, the minimum possible total sum is \( 404 \).

7. **Constructing a valid sequence**: To achieve a sum of \( 404 \), we can set \( a_i = 1 \) for \( i \equiv 1, 2, 3, 4, 5, 6 \pmod{10} \) and \( a_i = -1 \) for \( i \equiv 7, 8, 9, 0 \pmod{10} \) for \( i \leq 2010 \). Additionally, set \( a_{2011} = a_{2012} = 1 \). This configuration ensures that no 10 consecutive numbers sum to 0 and the total sum is \( 404 \).

8. **Achieving other sums**: Any even number larger than \( 404 \) can be attained by flipping the appropriate number of \( -1 \)'s to \( 1 \)'s. Thus, all even numbers \( S \) which satisfy \( 404 \leq |S| \leq 2012 \) can be obtained.

The final answer is \( \boxed{404, 406, 408, \ldots, 2012} \).