I will assume $N=\{1,2,3,\cdots\}$, otherwise we have the solution $f\equiv 0$. First choose $n$ with $f(n)$ minimal, then $n=1$ because otherwise $ff(n-1),ff(n+1)<f(n)$. Now consider $A=\{2,3,\cdots\}$ and choose $n\in A$ such that $f(n)=\min_{x\in A} f(x)$. Since $n\ge 2$ we have $ff(n-1),ff(n+1)<f(n)\implies f(n-1)=f(n+1)=1$, contradiction since $n+1>1$. Hence no such function.

The answer is:NO. We have f(n)>f(f(n-1))=>f(n+1)>f(f(n)). But,this implies f(n)=n (see the problem 6 of IMO 1977); but,this leads to: n=2n,false.. Thanks alexheinis for your solution!