There are at most $n$ differences between these numbers and by removing one of them, there are at most $n-1$ differences left, so it is impossible

As Tintarn has already mentioned, only $n \equiv 0 \pmod 4$ or $n \equiv 3 \pmod 4$ can work. Note that the differences are between $1$ and $n$. So when $n$ is a solution, then \[ |a_1-a_2|+|a_2-a_3|+\dots+|a_{n-1}-a_n|+|a_n-a_1| = \sum_{k=1}^n k = \frac{n(n+1)}{2}. \]Comparing both sides modulo $2$, the condition above follows. Now to construct solutions for both. For $n = 4k-1$ with $k \geq 2$ take \[ (a_1,a_2,\dots,a_{4k-1})=(1,4k,2,4k-1,\dots,k,3k+1,k+2,3k, k+3,\dots,2k+2,2k+1). \]A more formal definition of the sequenceDefine \[ a_{2i}=4k+1-i \quad \text{for } 1 \leq i \leq 2k-1, \qquad a_{2i-1} = \begin{cases} i \quad \ \ \ \ \ \text{for } 1 \leq i \leq k, \\ i+1 \quad \text{for } k < i \leq 2k. \end{cases} \] It's easy to verify that the above sequence satisfies the conditions of the problem. For $n = 4$ take $(a_1,a_2,a_3,a_4)=(1,3,2,5)$ and for $n=4k$ with $k \geq 2$ slightly modify the above construction. \[ (a_1,a_2,\dots,a_{4k-1},a_{4k})=(1,4k,2,4k-1,\dots,k,3k+1,k+2,3k, k+3,\dots,2k+2,2k+1, 4k+1) \]Once again, it's easy to verify that this sequence satisfies the conditions of the problem. Hence, we have shown that exactly $n \equiv 0, 3 \pmod 4$ work. $\hfill \square$

As Tintarn has already mentioned, only $n \equiv 0 \pmod 4$ or $n \equiv 3 \pmod 4$ can work. Note that the differences are between $1$ and $n$. So when $n$ is a solution, then \[ |a_1-a_2|+|a_2-a_3|+\dots+|a_{n-1}-a_n|+|a_n-a_1| = \sum_{k=1}^n k = \frac{n(n+1)}{2}. \]Comparing both sides modulo $2$, the condition above follows. Now to construct solutions for both. For $n = 4k-1$ with $k \geq 2$ take \[ (a_1,a_2,\dots,a_{4k-1})=(1,4k,2,4k-1,\dots,k,3k+1,k+2,3k, k+3,\dots,2k+2,2k+1). \]A more formal definition of the sequenceDefine \[ a_{2i}=4k+1-i \quad \text{for } 1 \leq i \leq 2k-1, \qquad a_{2i-1} = \begin{cases} i \quad \ \ \ \ \ \text{for } 1 \leq i \leq k, \\ i+1 \quad \text{for } k < i \leq 2k. \end{cases} \] It's easy to verify that the above sequence satisfies the conditions of the problem. For $n = 4$ take $(a_1,a_2,a_3,a_4)=(1,3,2,5)$ and for $n=4k$ with $k \geq 2$ slightly modify the above construction. \[ (a_1,a_2,\dots,a_{4k-1},a_{4k})=(1,4k,2,4k-1,\dots,k,3k+1,k+2,3k, k+3,\dots,2k+2,2k+1, 4k+1) \]Once again, it's easy to verify that this sequence satisfies the conditions of the problem. Hence, we have shown that exactly $n \equiv 0, 3 \pmod 4$ work. $\hfill \square$