Clearly the first number in the sequence is $0$. With an odd $n$ is impossible, since $0\equiv \sum_{i=0}^{n-1} i=\frac{n(n+1)}{2}$, since in the last formula the even factor is $n+1$ and so is divisible by $n$. With an even $n$ it works in many ways: the more simple is with the sequence $0, \ n-1, \ 2,\ n-3,\ 4,\ n-5,\ 6, \ \dots , n-4, \ 3, n-2,\ 1$.

This problem was proposed by Burii.

Problem is not original https://artofproblemsolving.com/community/q3h1365650p7504433

Riemann00 wrote: Clearly the first number in the sequence is $0$. With an odd $n$ is impossible, since $0\equiv \sum_{i=0}^{n-1} i=\frac{n(n+1)}{2}$, since in the last formula the even factor is $n+1$ and so is divisible by $n$. With an even $n$ it works in many ways: the more simple is with the sequence $0, \ n-1, \ 2,\ n-3,\ 4,\ n-5,\ 6, \ \dots , n-4, \ 3, n-2,\ 1$. Why does the first number has to be zero?

Arc_archer wrote: Riemann00 wrote: Clearly the first number in the sequence is $0$. With an odd $n$ is impossible, since $0\equiv \sum_{i=0}^{n-1} i=\frac{n(n+1)}{2}$, since in the last formula the even factor is $n+1$ and so is divisible by $n$. With an even $n$ it works in many ways: the more simple is with the sequence $0, \ n-1, \ 2,\ n-3,\ 4,\ n-5,\ 6, \ \dots , n-4, \ 3, n-2,\ 1$. Why does the first number has to be zero? Because if some other element is zero, say $x_k$, then $\sum_{i=0}^{k-1}x_i = \sum_{i=0}^{k}x_i$, therefore $x_0=0$

The answer is $n$ even, achieved by $0$, $1$, $n-2$, $3$, $n-4$, $5$, $\ldots$. Assume $n$ is odd. The key claim: Claim: $x_0=0$. Proof. Assume not, so $x_k=0$ for $k>0$. Then $x_0+\cdots+x_{k-1}=x_0+\cdots+x_k$, absurd. $\blacksquare$ However $x_0+x_1+\cdots+x_{n-1}=\frac{n(n-1)}2\equiv0=x_0\pmod n$, contradiction.