The answer is no. We shall prove by induction that $f(n) = 2n - 1$ and $g(n) = n$.

Notice that $f(i) \ge 1$ for each $i$. Then, if $1 = g(N) = \frac{1}{N}(f(1) + \cdots + f(N))$, it follows that $f(1) = \cdots = f(N) = 1$.

Now, we prove through induction on $n$. Assume that the statement holds for positive integers up to $n$. Since $g$ is surjective, assume that there is $m \ge n + 1$ such that $g(m) = n + 1$. Thus, at least one of the numbers $g(n + 1), \ldots, g(m)$ would be greater than or equal to $m$. Let there is some $k \in \{n + 1, \ldots, m\}$ such that $g(k) = m$ therefore,
$$g(m) = \frac{f(1) + \cdots + f(m)}{m} \ge \frac{f(1) + \cdots + f(k)}{m} = \frac{k \cdot g(k)}{m} \ge k \ge n+1$$
The equality case occurs and $k = m = n + 1$. Yielding $g(n + 1) = n + 1$ and $f(n + 1) = 2n - 1$, as desired. ■