Let $a_1,a_2,...,a_9$ be nine real numbers, not necessarily distinct, with average $m$. Let $A$ denote the number of triples $1 \le i < j < k \le 9$ for which $a_i + a_j + a_k \ge 3m$.  What is the minimum possible value of $A$?

[i]Proposed by Ray Li[/i]
1. **Given Information:**
   - We have nine real numbers \(a_1, a_2, \ldots, a_9\) with an average \(m\).
   - We need to find the minimum number of triples \((i, j, k)\) such that \(1 \le i < j < k \le 9\) and \(a_i + a_j + a_k \ge 3m\).

2. **Construction Example:**
   - Consider the numbers \(a_1 = a_2 = \cdots = a_8 = 0\) and \(a_9 = 187\).
   - The average \(m\) is given by:
     \[
     m = \frac{a_1 + a_2 + \cdots + a_9}{9} = \frac{0 + 0 + \cdots + 0 + 187}{9} = \frac{187}{9} \approx 20.78
     \]
   - For any triple \((i, j, k)\) to satisfy \(a_i + a_j + a_k \ge 3m\), at least one of \(i, j, k\) must be 9 because \(a_9\) is significantly larger than the others.
   - The number of such triples where one of the indices is 9 is:
     \[
     \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28
     \]

3. **Proof of Minimum Value:**
   - We need to show that \(A\) cannot be smaller than 28.
   - Consider all possible triples \((i, j, k)\) from the set \(\{1, 2, \ldots, 9\}\). The total number of such triples is:
     \[
     \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
     \]
   - We can partition these triples into groups of three sets \(A, B, C\) such that \(A \cap B = \emptyset\), \(B \cap C = \emptyset\), and \(C \cap A = \emptyset\). For example, \(A = \{1, 5, 6\}\), \(B = \{2, 8, 9\}\), \(C = \{3, 4, 7\}\).
   - Since \(A \cup B \cup C = \{1, 2, \ldots, 9\}\), the sum of the elements in each set must be at least \(3m\) to ensure the total sum is at least \(9m\).
   - Therefore, at least one of the sets \(A, B, C\) must satisfy \(a_i + a_j + a_k \ge 3m\).
   - This implies that at least \(\frac{1}{3} \binom{9}{3} = \frac{1}{3} \times 84 = 28\) triples must satisfy the condition.

Thus, the minimum possible value of \(A\) is 28.

The final answer is \(\boxed{28}\).