This is a strange problem... First, note that $ f(1) = 1$ is required, since $ f(n) = 1$ for some $ n$ but for all $ n\neq 1$, $ f(n) > 1$. Let $ S =\mathbb{N}-\{1\}$. We can see that on $ S$ the function $ n+(-1)^{n}$ is a bijection. So there is a unique value $ n$ in $ S$ for which $ f(n)$ can be $ 2$, and then there is a unique value $ n$ for which $ f(n)$ can be $ 3$, and so on. Because $ f$ is surjective and must take all values, it is required that $ f(n) = n+(-1)^{n}$ on $ S$. $ f(1) = 1$ and $ f(n) = n+(-1)^{n}$ for $ n > 1$.