We claim that the only solutions are:
1. $f(n) \equiv n + 1$. This works because
$$f^{f(m)}(n) + n = f^{m+1}(n) + n = m + 1 + 2n$$
and
$$f^m(n) + f(n) = m + n + n + 1 = m + 1 + 2n.$$
2. $f(n) \equiv n$. This works because
$$f^{f(m)}(n) + n = 2n = f^m(n) + f(n).$$
We now show these are the only solutions. Note that
$$f^{f^k(m)}(n) - f^{f^{k-1}(m)}(n) = f(n) - n.$$
Hence, by induction,
$$f^{f^k(m)}(n) = k(f(n) - n) + f^m(n).$$
This means that $f(n) \ge n$ for all positive integers $n$, else by choosing a sufficiently large $k$, $f^{f^k(m)}(n)$ will be negative. If there exists some positive integer $c$ such that $f(c) = c$, then
$$f^{f(c)}(n) = f^c(n) \Rightarrow f(n) = n \text{ for all positive integers } n.$$
Else, assume $f(n) \ne n$, so that $f(n) > n$ for all positive integers $n$. Pick $A$ such that $f(A) - A = t$ is minimal. For any positive integer $m$, consider the numbers $f^m(A), f^{m+1}(A), \dots, f^{f(m)}(A)$. This sequence of numbers is strictly increasing, with consecutive differences at least $t$. Hence,
$$t \le t(f(m) - m) \le f^{f(m)}(A) - f^m(A) = f(A) - A = t.$$
Thus, equality must hold everywhere, and in particular $f(m) = m + 1$ for all positive integers $m$, as desired.