IMO 4 : https://artofproblemsolving.com/community/c6h60761p366539

We can easily extend to $1987\rightarrow 2k-1(k\in \mathbb N)$. Let $A=\mathbb N-f(\mathbb N), B=f(\mathbb N)-f(f(\mathbb N))$.Then $A\cap B=\phi$, $A\cup B=\mathbb N-f(f(\mathbb N))=\{1,\dots ,2k-1\}$.Thus $|A|+|B|=|A\cup B|=2k-1$.On the other hand,obviously $f_A:A\rightarrow B$ is injective and surjective.Hence $|A|=|B|$,and then $|A|+|B|$ is even which is absurd.Therefore there is no such $f$.$\blacksquare$

http://www.artofproblemsolving.com/community/c6h389691p2165293 SPAIN 2000

@nicky-glass Thanks for teaching link.That problem can be solved by my solution.

I will interpret it with a graph. $f$ is clearly injective and since we have a function, every vertex has an out-degree of $1$ and in-degree of at most $1$. Assume that there is a cycle in this graph. It is obvious that this cycle doesn't have a length of $1$ or $2$. Assume the cycle is $a_1, a_2, ..., a_n$ where n is greater than $2$. After walking on a path of length $2$ we are basically adding $1987$ to our original number. We can't have $a_n=a_1+1987k=a_1$ for a positive $k$ and if $a_n=a_2+1987k=a_1$, then $f(a_n)=f(a_2+1987k)=a_1 \Rightarrow a_2=f(a_1)=f(f(a_2+1987k))=a_2+1987k+1987$ which is obviously wrong. This graph consists of non-intersecting directed infinite-length paths. Take the first two starting points of these paths. Any number smaller than $1988$ must be on these points since if they were on the third point or in a further point, there would be a nonpositive integer on this path coming before this number. However the total number of the first two starting points is even, meaning that we have to have some number, call it $x$, greater than $1987$ on these starting points on some path, call it $T$. We know that if $x-1987$ is on another path, call it $P$, there is $x$ on $P$ as well, contradicting the fact that paths do not cross each other. If $x$ and $x-1997$ is on $T$ together, there would be another $x$ on $T$, having an odd distance to the original $x$, contradiction.

Note that $f(f(f(n)))=f(n+1987)=f(n)+1987$ so $f(r+1987k)=f(r)+1987k$ for $r\in \{1,\dots,1987\}$ and integer $k\ge 0$. Consider the directed graph with vertices $\{1, \dots, 1987\}$ where we draw an edge from $a$ to $b$ iff $f(a)\equiv b\pmod{1987}$. As $f(f(n))\equiv n\pmod{1987}$, this graph must consist of cycles of length $2$ and length $1$. Since $1987$ is odd, one of these cycles must have length $1$. That is, $f(r)=r+1987k$ for some $r\in \{1,\dots,1987\}$ and integer $k\ge 0$. Note that $r+1987=f(f(r))=f(r+1987k)=f(r)+1987k=r+1987\cdot 2k$ so $k=\frac12$, which is clearly absurd. Therefore, there are no solutions for $f$.