Let $a_1,a_2,a_3,a_4$ be positive integers, with the property that it is impossible to assign them around a circle where all the neighbors are coprime. Let $i,j,k\in\{1,2,3,4\}$ with $i \neq j$, $j\neq k$, and $k\neq i $. Determine the maximum number of triples $(i,j,k)$ for which 
$$
({\rm gcd}(a_i,a_j))^2|a_k.
$$
To solve this problem, we need to determine the maximum number of triples \((i, j, k)\) for which \((\gcd(a_i, a_j))^2 \mid a_k\). We are given that it is impossible to arrange \(a_1, a_2, a_3, a_4\) around a circle such that all neighbors are coprime.

1. **Initial Assumptions and Setup**:
   - Assume \(a_1, a_2, a_3, a_4\) are positive integers.
   - We need to find the maximum number of triples \((i, j, k)\) such that \((\gcd(a_i, a_j))^2 \mid a_k\).

2. **Case 1: All Pairs Have \(\gcd > 1\)**:
   - Assume \((\gcd(a_i, a_j)) > 1\) for every \(i \neq j\).
   - Consider \(a_1 = p^{15}, a_2 = p^7, a_3 = p^3, a_4 = p\) for some prime \(p\).
   - Calculate \((\gcd(a_i, a_j))^2\) for each pair:
     \[
     \gcd(a_1, a_2) = p^7 \implies (\gcd(a_1, a_2))^2 = p^{14}
     \]
     \[
     \gcd(a_1, a_3) = p^3 \implies (\gcd(a_1, a_3))^2 = p^6
     \]
     \[
     \gcd(a_1, a_4) = p \implies (\gcd(a_1, a_4))^2 = p^2
     \]
     \[
     \gcd(a_2, a_3) = p^3 \implies (\gcd(a_2, a_3))^2 = p^6
     \]
     \[
     \gcd(a_2, a_4) = p \implies (\gcd(a_2, a_4))^2 = p^2
     \]
     \[
     \gcd(a_3, a_4) = p \implies (\gcd(a_3, a_4))^2 = p^2
     \]
   - Check divisibility:
     \[
     (\gcd(a_1, a_2))^2 = p^{14} \nmid a_3 = p^3 \quad \text{and} \quad p^{14} \nmid a_4 = p
     \]
     \[
     (\gcd(a_2, a_3))^2 = p^6 \nmid a_4 = p
     \]
     \[
     (\gcd(a_1, a_3))^2 = p^6 \nmid a_4 = p
     \]
   - This gives a total of 8 valid triples.

3. **Case 2: Existence of a Pair with \(\gcd = 1\)**:
   - Assume there exists a pair \((a_i, a_j)\) such that \(\gcd(a_i, a_j) = 1\).
   - Without loss of generality, assume \(\gcd(a_1, a_2) = 1\).
   - Place \(a_1\) and \(a_2\) on opposite sides of the circle.
   - At least one of \((a_1, a_3), (a_1, a_4), (a_2, a_3), (a_2, a_4)\) must have \(\gcd > 1\).
   - Assume \(\gcd(a_1, a_3) > 1\) and \(\gcd(a_1, a_2) > 1\).

4. **Contradiction and Maximum Number of Triples**:
   - If there are at least 9 triples, there must be a pair \((a_i, a_j)\) such that for \(k \in \{1, 2, 3, 4\} \setminus \{i, j\}\), \((\gcd(a_i, a_j))^2 \mid a_k\).
   - Let \(p\) be a prime dividing \(a_i\) and \(a_j\). Then \(p\) must divide \(a_1, a_2, a_3, a_4\).
   - Assume \(v_p(a_1) \geq v_p(a_2) \geq v_p(a_3) \geq v_p(a_4)\).
   - Note that \((\gcd(a_1, a_2))^2\) cannot divide \(a_3\) and \(a_4\), \((\gcd(a_1, a_3))^2\) cannot divide \(a_4\), and \((\gcd(a_2, a_3))^2\) cannot divide \(a_4\).
   - This results in a total of 8 valid triples.

Thus, the maximum number of triples \((i, j, k)\) for which \((\gcd(a_i, a_j))^2 \mid a_k\) is 8.

The final answer is \( \boxed{ 8 } \).