Suppose $c_i = a_i b_i = c_0 + it$ with $t \in \mathbb{N}$. Since $c_1, \dots, c_n$ are $n$ numbers in $\{n+1, \dots, 2n^2\}$, there are two with difference at most $\frac{2n^2 -(n+1)}{n-1} = 2n+1$, so $t \le 2n+1$. And if $t = 2n+1$, we need $c_n = 2n^2$, but then $c_{n-1} \le (n-1)(2n-1) < 2n^2 - (2n+1)$, so that indeed $t \le 2n$. Further, $c_n \ge n^2 + n$ and $c_1 \le 2n$ imply $t \ge n$. From $c_1 = c_0 + t \in \{n+1, \dots, 2n\}$, we then get $c_0 \in \{-n+1, \dots, n\}$. Now, we can locate $j$ so that $c_j = dt = c_0 + jt$ for some $d \in \{1, \dots, 2n\}.$ It follows that $t | c_0$, and so $t|a_i b_i$ for all $i$. If $t > n+1$, then it follows that $t|(t-1)k \implies t |k$ for some $k \in \{1, \dots, n\}$, absurd. If $t \le n+1 < 2n$, then it follows that $t | (t+1) k \implies t | k$ for some $k \ge \{1, \dots ,n\}$, again absurd. So there aren't any such arithmetic progressions.

Can you please explain the part after cj=dt