Can we put intengers $1,2,\cdots,12$ on a circle, number them $a_1,a_2,\cdots,a_{12}$ in order. For any $1\leq i<j\leq12$, $|a_i-a_j|\neq|i-j|$?
1. **Restate the problem in terms of permutations:**
   We need to find a permutation $\{a_i\}$ of the set $\{1, 2, \ldots, 12\}$ such that for any $1 \leq i < j \leq 12$, the absolute difference $|a_i - a_j|$ is not equal to the absolute difference $|i - j|$.

2. **Reformulate the problem using chessboard analogy:**
   This problem can be rephrased as finding a permutation $\{a_i\}$ such that no two $(a_i + i)$'s are equal and no two $(a_i - i)$'s are equal. This is analogous to placing 12 queens on a $12 \times 12$ chessboard such that no two queens attack each other.

3. **Find a valid permutation:**
   After experimenting with different permutations, we find that the permutation $(1, 6, 8, 10, 7, 4, 2, 9, 12, 3, 5, 11)$ satisfies the conditions. Let's verify this permutation:

   - For $(a_i + i)$:
     \[
     \begin{aligned}
     &a_1 + 1 = 1 + 1 = 2, \\
     &a_2 + 2 = 6 + 2 = 8, \\
     &a_3 + 3 = 8 + 3 = 11, \\
     &a_4 + 4 = 10 + 4 = 14, \\
     &a_5 + 5 = 7 + 5 = 12, \\
     &a_6 + 6 = 4 + 6 = 10, \\
     &a_7 + 7 = 2 + 7 = 9, \\
     &a_8 + 8 = 9 + 8 = 17, \\
     &a_9 + 9 = 12 + 9 = 21, \\
     &a_{10} + 10 = 3 + 10 = 13, \\
     &a_{11} + 11 = 5 + 11 = 16, \\
     &a_{12} + 12 = 11 + 12 = 23.
     \end{aligned}
     \]
     All values are distinct.

   - For $(a_i - i)$:
     \[
     \begin{aligned}
     &a_1 - 1 = 1 - 1 = 0, \\
     &a_2 - 2 = 6 - 2 = 4, \\
     &a_3 - 3 = 8 - 3 = 5, \\
     &a_4 - 4 = 10 - 4 = 6, \\
     &a_5 - 5 = 7 - 5 = 2, \\
     &a_6 - 6 = 4 - 6 = -2, \\
     &a_7 - 7 = 2 - 7 = -5, \\
     &a_8 - 8 = 9 - 8 = 1, \\
     &a_9 - 9 = 12 - 9 = 3, \\
     &a_{10} - 10 = 3 - 10 = -7, \\
     &a_{11} - 11 = 5 - 11 = -6, \\
     &a_{12} - 12 = 11 - 12 = -1.
     \end{aligned}
     \]
     All values are distinct.

4. **Conclusion:**
   Since the permutation $(1, 6, 8, 10, 7, 4, 2, 9, 12, 3, 5, 11)$ satisfies both conditions, we conclude that it is possible to arrange the integers $1, 2, \ldots, 12$ on a circle such that $|a_i - a_j| \neq |i - j|$ for any $1 \leq i < j \leq 12$.

The final answer is $\boxed{\text{yes}}$