Problem 2490

Suppose that a function $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfies (1). For arbitrary positive integers $m>n$, by (1) we have
$$f(m)=f(n+(m-n)) \geq f(n)+f(f(m-n))-1 \geq f(n),$$
so $f$ is nondecreasing.

Function $f \equiv 1$ is an obvious solution. To find other solutions, assume that $f \not \equiv 1$ and take the smallest $a \in \mathbb{N}$ such that $f(a)>1$. Then $f(b) \geq f(a)>1$ for all integer $b \geq a$.

Suppose that $f(n)>n$ for some $n \in \mathbb{N}$. Then we have
$$f(f(n))=f((f(n)-n)+n) \geq f(f(n)-n)+f(f(n))-1$$
so $f(f(n)-n) \leq 1$ and hence $f(n)-n<a$. Then there exists a maximal value of the expression $f(n)-n$; denote this value by $c$, and let $f(k)-k=c \geq 1$. Applying the monotonicity together with (1), we get
$$\begin{aligned} 2 k+c \geq f(2 k)=f(k+k) & \geq f(k)+f(f(k))-1 \\ & \geq f(k)+f(k)-1=2(k+c)-1=2 k+(2 c-1), \end{aligned}$$
hence $c \leq 1$ and $f(n) \leq n+1$ for all $n \in \mathbb{N}$. In particular, $f(2007) \leq 2008$.

Now we present a family of examples showing that all values from 1 to 2008 can be realized. Let
$$f_{j}(n)=\max \{1, n+j-2007\} \quad \text{ for } j=1,2, \ldots, 2007 ; \quad f_{2008}(n)= \begin{cases}n, & 2007 \nmid n, \\ n+1, & 2007 \mid n .\end{cases}$$
We show that these functions satisfy the condition (1) and clearly $f_{j}(2007)=j$.

To check the condition (1) for the function $f_{j}(j \leq 2007)$, note first that $f_{j}$ is nondecreasing and $f_{j}(n) \leq n$, hence $f_{j}\left(f_{j}(n)\right) \leq f_{j}(n) \leq n$ for all $n \in \mathbb{N}$. Now, if $f_{j}(m)=1$, then the inequality (1) is clear since $f_{j}(m+n) \geq f_{j}(n) \geq f_{j}\left(f_{j}(n)\right)=f_{j}(m)+f_{j}\left(f_{j}(n)\right)-1$. Otherwise,
$$f_{j}(m)+f_{j}\left(f_{j}(n)\right)-1 \leq(m+j-2007)+n=(m+n)+j-2007=f_{j}(m+n) .$$
In the case $j=2008$, clearly $n+1 \geq f_{2008}(n) \geq n$ for all $n \in \mathbb{N}$; moreover, $n+1 \geq f_{2008}\left(f_{2008}(n)\right)$ as well. Actually, the latter is trivial if $f_{2008}(n)=n$; otherwise, $f_{2008}(n)=n+1$, which implies $2007 \nmid n+1$ and hence $n+1=f_{2008}(n+1)=f_{2008}\left(f_{2008}(n)\right)$.

So, if $2007 \mid m+n$, then
$$f_{2008}(m+n)=m+n+1=(m+1)+(n+1)-1 \geq f_{2008}(m)+f_{2008}\left(f_{2008}(n)\right)-1 .$$
Otherwise, $2007 \nmid m+n$, hence $2007 \nmid m$ or $2007 \nmid n$. In the former case we have $f_{2008}(m)=m$, while in the latter one $f_{2008}\left(f_{2008}(n)\right)=f_{2008}(n)=n$, providing
$$f_{2008}(m)+f_{2008}\left(f_{2008}(n)\right)-1 \leq(m+n+1)-1=f_{2008}(m+n) \text{.}$$