The answer is all even $n$. For $n=2k$ we can take $f(i)=f(i+1)=i$ and $g(i)=g(i+1)=i+1$, for $i=1,3\dots,2k-1$. Assume now for contradiction that there are such functions $f,g$ such that the conditions hold for some odd $n=2k+1$, then by PHP we know that w.l.o.g $f(g(x))=x$ happens at least $k+1$ times and w.l.o.g $f(g(i))=i$ for $1\leq i\leq k+1$, then again by PHP there is a $j (1\leq j \leq k+1)$ such that $g(j)=\ell$ where $1\leq \ell \leq k+1$, now we have $f(g(j))=j$ which implies $f(\ell)=j$ and we have $g(f(\ell))=\ell$ which is not possible.