Find all triples $(a,b,c)$ of positive integers such that the product of any two of them when divided by the third leaves the remainder $1$.

1. We start with the given conditions:
   \[
   a \mid bc - 1, \quad b \mid ac - 1, \quad c \mid ab - 1
   \]
   where \(a, b, c\) are positive integers greater than 1.

2. Without loss of generality, assume \(a \leq b \leq c\).

3. Let \(d = \gcd(a, b)\). Since \(d \mid a\) and \(d \mid b\), it follows that \(d \mid \gcd(a, b)\). From the condition \(a \mid bc - 1\), we have:
   \[
   d \mid bc - 1
   \]
   Since \(d \mid b\), it follows that \(d \mid 1\). Therefore, \(d = 1\). Similarly, we can show that \(\gcd(a, c) = 1\) and \(\gcd(b, c) = 1\). Thus, \(a, b, c\) are pairwise coprime.

4. Using the conditions, we can derive:
   \[
   abc \mid (bc - 1)(ac - 1)(ab - 1)
   \]
   Expanding the right-hand side, we get:
   \[
   (bc - 1)(ac - 1)(ab - 1) = (abc)^2 - abc(a + b + c) + (ab + ac + bc) - 1
   \]
   Therefore:
   \[
   abc \mid (abc)^2 - abc(a + b + c) + (ab + ac + bc) - 1
   \]
   Dividing both sides by \(abc\), we obtain:
   \[
   1 \leq \frac{ab + ac + bc - 1}{abc} = F(a, b, c) \in \mathbb{N}
   \]

5. Since \(a \leq b \leq c\), we have:
   \[
   \frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}
   \]
   Therefore:
   \[
   1 \leq F(a, b, c) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{1}{abc} < \frac{3}{a}
   \]
   This implies:
   \[
   \frac{3}{a} > 1 \implies a < 3
   \]
   Since \(a\) is a positive integer greater than 1, we have \(a = 2\).

6. Substituting \(a = 2\) into the inequality:
   \[
   1 \leq F(2, b, c) < \frac{3}{2}
   \]
   This implies:
   \[
   F(2, b, c) = 1
   \]
   Therefore:
   \[
   \frac{2b + 2c + bc - 1}{2bc} = 1
   \]
   Simplifying, we get:
   \[
   bc - 2b - 2c + 1 = 0
   \]
   Rearranging, we have:
   \[
   (b - 2)(c - 2) = 3
   \]

7. Since \(2 = a < b < c\), the only pairs \((b - 2, c - 2)\) that satisfy the equation are \((1, 3)\) and \((3, 1)\). Thus:
   \[
   b - 2 = 1 \implies b = 3, \quad c - 2 = 3 \implies c = 5
   \]
   or
   \[
   b - 2 = 3 \implies b = 5, \quad c - 2 = 1 \implies c = 3
   \]
   However, since \(b < c\), the only valid solution is \(b = 3\) and \(c = 5\).

8. Therefore, the only positive integer solutions \((a, b, c)\) are \((2, 3, 5)\) and its permutations.

The final answer is \( \boxed{ (2, 3, 5) } \) and its permutations.