Find all natural numbers $n$ greater than $2$ such that there exist $n$ natural numbers $a_{1},a_{2},\ldots,a_{n}$ such that they are not all equal, and the sequence $a_{1}a_{2},a_{2}a_{3},\ldots,a_{n}a_{1}$ forms an arithmetic progression with nonzero common difference.
1. **Assume the common difference of the arithmetic progression (AP) is positive.** If the common difference is negative, we can consider the AP in reverse order: \(a_1a_2, a_2a_3, \ldots, a_{n-1}a_n, a_na_1\). This does not change the nature of the problem.

2. **Establish the inequality \(a_1a_2 < a_2a_3\).** This implies \(a_1 < a_3\).

3. **Formulate the condition for the sequence to be an AP.** The sequence \(a_1a_2, a_2a_3, \ldots, a_na_1\) forms an AP if and only if for all \(1 \leq i \leq n-2\),
   \[
   \frac{a_i}{a_{i+2}} + \frac{a_{i+3}}{a_{i+1}} = 2 \quad \forall \, 1 \leq i \leq n-2 \tag{1}
   \]

4. **Express \(a_1\) in terms of \(a_3\).** Since \(a_1 < a_3\), there exists \(t \in \mathbb{Q}_{>0}\) such that
   \[
   \frac{a_1}{a_3} = \frac{t}{t+1}
   \]

5. **Use induction to generalize the ratio.** By induction, we can show that equation (1) is equivalent to
   \[
   \frac{a_i}{a_{i+2}} = \frac{t+i-1}{t+i} \quad \forall \, 1 \leq i \leq n-1 \tag{2}
   \]

6. **Analyze the product of ratios for even \(n\).** When \(n\) is even, we have
   \[
   1 = \frac{a_1}{a_3} \cdot \frac{a_3}{a_5} \cdots \frac{a_{n-1}}{a_1} = \frac{t}{t+1} \cdot \frac{t+2}{t+3} \cdots \frac{t+n-2}{t+n-1}
   \]
   This product cannot equal 1 for any \(t\), as the terms do not cancel out to 1.

7. **Analyze the product of ratios for odd \(n\).** When \(n\) is odd, choosing \(a_1\) randomly determines all other \(a_i\) uniquely. This means equation (2) can be satisfied for some \(a_1, \ldots, a_n \in \mathbb{Q}_{>0}\). We can then multiply all \(a_1, \ldots, a_n\) by the least common multiple (LCM) of their denominators to ensure they are natural numbers.

8. **Conclusion.** The sequence \(a_1a_2, a_2a_3, \ldots, a_na_1\) forms an arithmetic progression with nonzero common difference if and only if \(n\) is odd.

\(\blacksquare\)

The final answer is \( \boxed{ n } \) is odd.