Note that if the product of any two distinct members of {1,16,27} is increased by 9, the result is the perfect square of an integer. Find the unique positive integer $n$ for which $n+9,16n+9,27n+9$ are also perfect squares.
Given the set \(\{1, 16, 27\}\), we know that the product of any two distinct members of this set, when increased by 9, results in a perfect square. We need to find the unique positive integer \(n\) such that \(n+9\), \(16n+9\), and \(27n+9\) are also perfect squares.

1. Assume \(n = k^2 - 9\) for some integer \(k\). Then:
   \[
   n + 9 = k^2
   \]
   This implies:
   \[
   16n + 9 = 16(k^2 - 9) + 9 = 16k^2 - 144 + 9 = 16k^2 - 135
   \]
   Let \(l^2 = 16k^2 - 135\). This can be rewritten as:
   \[
   (4k - l)(4k + l) = 135
   \]

2. Let \(d = 4k - l\). Then:
   \[
   d(d + 2l) = 135
   \]
   Since \(d\) must be a divisor of 135 and \(d < \sqrt{135} \approx 11.62\), we consider the divisors of 135 that are less than 12: \(1, 3, 5, 9\).

3. Check each possible value of \(d\):

   - For \(d = 1\):
     \[
     1(1 + 2l) = 135 \implies 1 + 2l = 135 \implies 2l = 134 \implies l = 67
     \]
     \[
     4k - 67 = 1 \implies 4k = 68 \implies k = 17
     \]
     \[
     n = k^2 - 9 = 17^2 - 9 = 289 - 9 = 280
     \]
     Verify:
     \[
     280 + 9 = 289 = 17^2
     \]
     \[
     16 \cdot 280 + 9 = 4480 + 9 = 4489 = 67^2
     \]
     \[
     27 \cdot 280 + 9 = 7560 + 9 = 7569 = 87^2
     \]
     Thus, \(n = 280\) is a valid solution.

   - For \(d = 3\):
     \[
     3(3 + 2l) = 135 \implies 3 + 6l = 135 \implies 6l = 132 \implies l = 22
     \]
     \[
     4k - 22 = 3 \implies 4k = 25 \implies k = 6.25 \quad (\text{not an integer})
     \]
     This does not yield a valid \(k\).

   - For \(d = 5\):
     \[
     5(5 + 2l) = 135 \implies 5 + 10l = 135 \implies 10l = 130 \implies l = 13
     \]
     \[
     4k - 13 = 5 \implies 4k = 18 \implies k = 4.5 \quad (\text{not an integer})
     \]
     This does not yield a valid \(k\).

   - For \(d = 9\):
     \[
     9(9 + 2l) = 135 \implies 9 + 18l = 135 \implies 18l = 126 \implies l = 7
     \]
     \[
     4k - 7 = 9 \implies 4k = 16 \implies k = 4
     \]
     \[
     n = k^2 - 9 = 4^2 - 9 = 16 - 9 = 7
     \]
     Verify:
     \[
     27 \cdot 7 + 9 = 189 + 9 = 198 \quad (\text{not a perfect square})
     \]
     Thus, \(n = 7\) is not a valid solution.

4. Since \(n = 280\) is the only valid solution, we conclude that the unique positive integer \(n\) is 280.

The final answer is \(\boxed{280}\)