Any odd $n \ge 3$ works, and all even $n$ fail. To see that even $n$ fail, note that the product of every even numbered term in the AP is equal to the product of every odd numbered term in the AP. If the common difference is positive, it is clear to see the second is larger than the first, and if the common difference is negative, then it is clear to see that the first is larger than the second, both leading to contradictions. For $n$ odd, the following works: consider the positive rational sequence $b_i$ with $b_1=1$, $b_2=1$, and $b_ib_{i-1}=n-1$ for $1 \le i \le n$. Now, consider the rational sequence $c_i$, with $c_{2i}=b_{2i}$, and $c_{2i+1}=r\cdot b_{2i+1}$, where $r=\frac{n}{b_nb_1}$. Then $c_1c_2,c_2c_3,...,c_nc_1$ forms a rational arithmetic progression with common difference $r \ne 0$. Now, $a_i=k\cdot c_i$ works, where $k$ is chosen to be the least common multiple of all the denominators of the $c_i$ written in lowest form. Cheers, Rofler

The problem is proposed by Morteza Saghafian.

First of all $a_1=a_{n-1}+\frac {d}{a_n}$.Now also $a_n=a_2+\frac {(n-1)d}{a_1}$ and $a_{n-1}=\frac {(n-1)d+a_1a_2}{a_1a_2+(n-1)d}$. Now so we get $\frac {a_1a_2+(n-2)d+da_1}{a_1a_2+(n-1)d}=a_1$. So now we must have $d(a_1-1)=0\implies x_1=0$.Now so if $n$ is even then $a_n$ must be of form $a_2+df(a_2)$ and from reverse direction we get $a_n=a_2+(n-1)d$ thus we have $f(a_2)=n-1$. But it's easy to $a_{odd}>1$ so certainly $f(a_2)<1$. Thus $n$ must be odd. Now it's easy to show that's not possible for $n=3$.Now for a construction choose $y_iy_{i+1}=ki$ for all $1<i\leq n$ where $k$ is a constant. Now we can choose up to $y_{n-1}$ without any problem, but when it comes $y_n$ we're getting like $y_n=f(n)y_1=\frac {kn}{y_1}$ so taking $k$ such that $\frac {kn}{f(n)}$ is square of a rational we're done to find a rational sequence of AP. Now now multiplying all by their LCM of the denominators we're done.

When $n$ is even,say of form $2k$ we have $a_2(a_1-a_3)=a_4(a_3-a_5)=\cdots=a_n(a_{n-1}-a_1)=x$ where $x$ is obviously the negative of the common difference. Then it is easy to note that $x\sum_{i=1}^{\frac{n}{2}}{\frac{1}{a_{2i}}}=0$.This implies $x=0$ which cannot hold by the question. For odd $n$ my construction is coincidentally same with Rofler's.(Basically there's a typo in Rofler's solution- it will be $b_ib_{i+1}=i$

Saying this thing is an arithmetic progression is the same as saying a_i*a_(i+1) - a_i*a_(i-1) = a_(i+2)*a_(i+1) - a_i*a_(i+1) for every i. Rewrite as 2= a_(i-1)/a_(i+1) + a_(i+2)/a_i so sum up all expressions for all possible values of i and use the fact that a/b + b/a >=2 with equality case only when a=b to deduce a_i=a_(i+2) for all i.

Mikeglicker wrote: Saying this thing is an arithmetic progression is the same as saying a_i*a_(i+1) - a_i*a_(i-1) = a_(i+2)*a_(i+1) - a_i*a_(i+1) for every i. Rewrite as 2= a_(i-1)/a_(i+1) + a_(i+2)/a_i so sum up all expressions for all possible values of i and use the fact that a/b + b/a >=2 with equality case only when a=b to deduce a_i=a_(i+2) for all i. it's wrong...

Here's a different solution which is easily motivated: Wlog assume common difference of AP is positive, as otherwise we can consider the AP $a_1a_n,a_{n}a_{n-1},a_{n-1}a_{n-2},\ldots,a_2a_1$. So $a_1a_2 < a_2a_3 \implies a_1 < a_3$. Note that the whole sequence will be a AP iff $a_ia_{i+1},a_{i+1}a_{i+2},a_{i+2}a_{i+3}$ is a AP for all $1 \le i \le n$ which is equivalent to $$\frac{a_i}{a_{i+2}} + \frac{a_{i+3}}{a_{i+1}} = 2 ~~ \forall ~ 1 \le i \le n-2 \qquad \qquad (1)$$Now since $a_1 < a_3$, so $\exists ~ t \in \mathbb Q_{>0}$ such that $$\frac{a_1}{a_3} = \frac{t}{t+1}$$Then by induction, we easily get that $(1)$ is equivalent to $$ \frac{a_i}{a_{i+2}} = \frac{t+i-1}{t+i} ~~ \forall ~ 1 \le i \le n-1 \qquad \qquad (2) $$We want to prove $(2)$ can be satisfied for some $a_1,\ldots,a_n \in \mathbb Z_{>0}$ and $t \in \mathbb Q_{>0}$ iff $n$ is odd. Note that when $n$ is even, then $$1 = \frac{a_1}{a_3} \cdot \frac{a_3}{a_5} \cdots \frac{a_{n-1}}{a_1} = \frac{t}{t+1} \cdot \frac{t+2}{t+3} \cdots \frac{t - n - 2}{t- n - 1} $$But clearly the above cannot be satisfied for any $t$. Now when $n$ is odd, then if we just choose $a_1$ randomly then everything just gets defined uniquely. This means $(2)$ can be satisfied for some $a_1,\ldots,a_n \in \mathbb Q_{>0}$, but then we just multiply all of $a_1,\ldots,a_n$ by the lcm of their denominators. $\blacksquare$