 Let $A_n$ be the number of arranged n-tuples of natural numbers $(a_1,a_2…a_n)$, such that
$\frac{1}{a_1} +\frac{1}{a_2} +...+\frac{1}{a_n} =1$.
Find the parity of $A_{68}$.

1. **Define the problem in terms of multisets:**
   Let $\mathcal{A}$ be the family of all multisets $A$ of exactly $n$ natural numbers for which 
   \[
   \sum_{a \in A} \frac{1}{a} = 1.
   \]
   For each $A \in \mathcal{A}$, let $p_A$ be the number of all distinct permutations (ordered $n$-tuples) of elements of $A$. Clearly, 
   \[
   A_n = \sum_{A \in \mathcal{A}} p_A.
   \]

2. **Consider the case when $n$ is even:**
   Let $n$ be even. We need to determine the parity (odd or even) of $A_n$. 

3. **Analyze the permutations:**
   Each $p_A$ is the number of distinct permutations of the multiset $A$. If $A$ consists of $n$ elements, then $p_A$ is the number of ways to arrange these $n$ elements. 

4. **Identify the special case:**
   The only case where $p_A$ is odd is when $A$ consists of $n$ equal elements. In this case, $p_A = 1$ because there is only one way to arrange $n$ identical elements. 

5. **Determine the specific multiset:**
   There is only one such multiset $A$ in $\mathcal{A}$, namely when all $n$ elements of $A$ are equal to $\frac{1}{n}$. For example, if $n = 68$, then the multiset $A$ would be $\left\{\frac{1}{68}, \frac{1}{68}, \ldots, \frac{1}{68}\right\}$.

6. **Summarize the parity:**
   Since there is exactly one such multiset where $p_A = 1$ and all other multisets have $p_A$ even, the sum $A_n$ will be odd. This is because the sum of an odd number (1) and any number of even numbers is odd.

\[
A_n = \sum_{A \in \mathcal{A}} p_A \equiv 1 \pmod{2}.
\]

The final answer is $\boxed{\text{odd}}$.