Determine all positive integers $n$ for which there exist positive integers $a_1,a_2, ..., a_n$ 

with $a_1 + 2a_2 + 3a_3 +... + na_n = 6n$ and $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+ ... +\frac{n}{a_n}= 2 + \frac1n$
1. **Applying the Cauchy-Schwarz Inequality:**

   We start by applying the Cauchy-Schwarz inequality to the given sums. The Cauchy-Schwarz inequality states that for any sequences of real numbers \(x_i\) and \(y_i\), we have:
   \[
   \left( \sum_{i=1}^n x_i y_i \right)^2 \leq \left( \sum_{i=1}^n x_i^2 \right) \left( \sum_{i=1}^n y_i^2 \right)
   \]
   In our case, let \(x_i = \sqrt{i a_i}\) and \(y_i = \frac{1}{\sqrt{i a_i}}\). Then:
   \[
   \left( \sum_{i=1}^n i \right)^2 \leq \left( \sum_{i=1}^n i a_i \right) \left( \sum_{i=1}^n \frac{i}{a_i} \right)
   \]
   Substituting the given conditions:
   \[
   \left( \frac{n(n+1)}{2} \right)^2 \leq (6n) \left( 2 + \frac{1}{n} \right)
   \]

2. **Simplifying the Inequality:**

   Simplify the right-hand side:
   \[
   \left( \frac{n(n+1)}{2} \right)^2 \leq 6n \left( 2 + \frac{1}{n} \right)
   \]
   \[
   \left( \frac{n(n+1)}{2} \right)^2 \leq 6n \left( 2 + \frac{1}{n} \right)
   \]
   \[
   \left( \frac{n(n+1)}{2} \right)^2 \leq 12n + 6
   \]
   \[
   \frac{n^2(n+1)^2}{4} \leq 12n + 6
   \]
   \[
   n^4 + 2n^3 + n^2 \leq 48n + 24
   \]
   \[
   n^4 + 2n^3 + n^2 - 48n - 24 \leq 0
   \]

3. **Checking Small Values of \(n\):**

   We need to check small values of \(n\) to see if they satisfy the inequality and the original conditions.

   - For \(n = 1\):
     \[
     a_1 = 6
     \]
     \[
     a_1 = 6 \implies \frac{1}{a_1} = \frac{1}{6} \neq 2 + \frac{1}{1} = 3
     \]
     So, \(n = 1\) does not work.

   - For \(n = 2\):
     \[
     a_1 + 2a_2 = 12
     \]
     \[
     \frac{1}{a_1} + \frac{2}{a_2} = 2 + \frac{1}{2} = \frac{5}{2}
     \]
     Solving these equations, we find that there are no positive integer solutions for \(a_1\) and \(a_2\) that satisfy both conditions. So, \(n = 2\) does not work.

   - For \(n = 3\):
     \[
     a_1 + 2a_2 + 3a_3 = 18
     \]
     \[
     \frac{1}{a_1} + \frac{2}{a_2} + \frac{3}{a_3} = 2 + \frac{1}{3} = \frac{7}{3}
     \]
     We can check the values \(a_1 = 6\), \(a_2 = 3\), and \(a_3 = 2\):
     \[
     6 + 2 \cdot 3 + 3 \cdot 2 = 6 + 6 + 6 = 18
     \]
     \[
     \frac{1}{6} + \frac{2}{3} + \frac{3}{2} = \frac{1}{6} + \frac{4}{6} + \frac{9}{6} = \frac{14}{6} = \frac{7}{3}
     \]
     So, \(n = 3\) works.

4. **Conclusion:**

   From the above steps, we see that \(n = 3\) is the only value that satisfies both conditions.

The final answer is \( \boxed{ n = 3 } \)