Solution:

Suppose that such a permutation exists. Since $\{p_{i}+i \mid 1 \leq i \leq n\}$ is a complete system of residues modulo $n$, we have $\sum_{k=1}^{n} k \equiv \sum_{i=1}^{n}(p_{i}+i) \equiv \sum_{i=1}^{n} i+\sum_{i=1}^{n} p_{i} \equiv 2 \sum_{k=1}^{n} k\ (\bmod\ n)$, hence $\sum_{k=1}^{n} k=\frac{n(n+1)}{2} \equiv 0\ (\bmod\ n)$, from which it follows that $2 \nmid n$.

Moreover, we have $2 \sum_{k=1}^{n} k^{2} \equiv \sum_{k=1}^{n}\left((p_{i}+i)^{2}+(p_{i}-i)^{2}\right) \equiv \sum_{k=1}^{n}\left(2 p_{i}^{2}+2 i^{2}\right) \equiv 4 \sum_{k=1}^{n} k^{2}$, from which $2 \sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{3} \equiv 0\ (\bmod\ n)$, hence $3 \nmid n$.

Therefore, we must have $(n, 6)=1$.

On the other hand, if $(n, 6)=1$ and $p_{i} \equiv 2 i\ (\bmod\ n),\ p_{i} \in\{1, \ldots, n\}$, then $(p_{1}, p_{2}, \ldots, p_{n})$ is a permutation of the set $\{1, \ldots, n\}$ and satisfies the conditions, since $\{p_{i}+i \mid 1 \leq i \leq n\} \equiv\{3 i \mid 1 \leq i \leq n\}$ and $\{p_{i}-i \mid 1 \leq i \leq n\} \equiv\{i \mid 1 \leq i \leq n\}$ $(\bmod\ n)$ are complete systems of residues modulo $n$.