Find the smallest positive integer $k$ such that \[(16a^2 + 36b^2 + 81c^2)(81a^2 + 36b^2 + 16c^2) < k(a^2 + b^2 + c^2)^2,\] for some ordered triple of positive integers $(a,b,c)$.
1. **Apply the Cauchy-Schwarz Inequality:**
   The Cauchy-Schwarz Inequality states that for any sequences of real numbers \(x_1, x_2, \ldots, x_n\) and \(y_1, y_2, \ldots, y_n\),
   \[
   \left( \sum_{i=1}^n x_i^2 \right) \left( \sum_{i=1}^n y_i^2 \right) \geq \left( \sum_{i=1}^n x_i y_i \right)^2.
   \]
   In our case, we can apply this inequality to the sequences \((4a, 6b, 9c)\) and \((9a, 6b, 4c)\):
   \[
   (16a^2 + 36b^2 + 81c^2)(81a^2 + 36b^2 + 16c^2) \geq (36a^2 + 36b^2 + 36c^2)^2.
   \]

2. **Simplify the Right-Hand Side:**
   \[
   (36a^2 + 36b^2 + 36c^2)^2 = 36^2(a^2 + b^2 + c^2)^2 = 1296(a^2 + b^2 + c^2)^2.
   \]
   Therefore, we have:
   \[
   (16a^2 + 36b^2 + 81c^2)(81a^2 + 36b^2 + 16c^2) \geq 1296(a^2 + b^2 + c^2)^2.
   \]
   This implies that \(k > 1296\).

3. **Check if \(k = 1297\) works:**
   We need to show that:
   \[
   (16a^2 + 36b^2 + 81c^2)(81a^2 + 36b^2 + 16c^2) < 1297(a^2 + b^2 + c^2)^2
   \]
   for some positive integers \(a, b, c\). Let's test with \(a = 1\), \(c = 1\), and vary \(b\).

4. **Substitute \(a = 1\) and \(c = 1\):**
   \[
   (16 \cdot 1^2 + 36b^2 + 81 \cdot 1^2)(81 \cdot 1^2 + 36b^2 + 16 \cdot 1^2) < 1297(1^2 + b^2 + 1^2)^2.
   \]
   Simplify the left-hand side:
   \[
   (16 + 36b^2 + 81)(81 + 36b^2 + 16) = (97 + 36b^2)(97 + 36b^2).
   \]
   Simplify the right-hand side:
   \[
   1297(1 + b^2 + 1)^2 = 1297(2 + b^2)^2.
   \]

5. **Compare the expressions:**
   \[
   (97 + 36b^2)^2 < 1297(2 + b^2)^2.
   \]
   Let \(x = b^2\). Then we need to show:
   \[
   (97 + 36x)^2 < 1297(2 + x)^2.
   \]
   Expand both sides:
   \[
   9409 + 2 \cdot 97 \cdot 36x + 36^2x^2 < 1297(4 + 4x + x^2).
   \]
   Simplify:
   \[
   9409 + 6984x + 1296x^2 < 5188 + 5188x + 1297x^2.
   \]
   Combine like terms:
   \[
   4221 + 1796x < x^2.
   \]
   This inequality holds for sufficiently large \(x\) (i.e., \(b\)). Therefore, \(k = 1297\) works.

\(\blacksquare\)

The final answer is \( \boxed{ k = 1297 } \).