Consider those functions \( f: \mathbb{N} \rightarrow \mathbb{N} \) which satisfy the condition 
\[ f(m+n) \geq f(m) + f(f(n)) - 1 \]
for all \( m, n \in \mathbb{N} \). Find all possible values of \( f(2007) \).  \((\mathbb{N}\) denotes the set of all positive integers.)  (Bulgaria)  Answer. 1, 2, .., 2008.
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 $f(n)-n \leq a$. 

$$ h_{j}(n)=\max \left\{1,\left\lfloor\frac{j n}{2007}\right\rfloor\right\} $$

Also the example for $j=2008$ can be generalized. In particular, choosing a divisor $d>1$ of 2007, one can set 

$$ f_{2008, d}(n)= \begin{cases}n, & d \nmid n \\ n+1, & d \mid n\end{cases} $$