67th NMO Selection Tests for BMO and IMO

If $n$ is odd, the required functions are the constant function $f_0\equiv 2$ along with the $n$ functions $f_i:\{1,\dots ,n\}\to {\mathbb{N}}^{\ast }$, $$f_i(j)=\{\begin{array}{ll}n+1,& \text{if }j=i,\\ 1,& \text{if }j\ne i,\end{array}i=1,\dots ,n;$$

notice that $f_0=f_1$ if $n=1$. If $n$ is even, $f_0$ is to be removed from the list, by (2). All these functions plainly satisfy the conditions in the statement. Labelling the positive integers in the list $f(1),\dots ,f(n)$ increasingly, $a_1\le \cdots \le a_n$, the problem amounts to determining all lists of positive integers $a_1\le \cdots \le a_n$ such that $(1^{\prime }){\sum }_{k=1}^n{a}_k=2n$ and $(2^{\prime }){\sum }_{k\in K}{a}_k=n$ for no subset $K$ of the first $n$ positive integers. Notice that $a_n\le n+1$, by $(1^{\prime })$, and $a_n\ne n$, by $(2^{\prime })$. With reference again to $(1^{\prime })$, notice further that if $a_n=n+1$, then the other $a_k$ are all 1, and if $a_n=2$, then so are the other $a_k$. If $n$ is even, $(2^{\prime })$ rules out the latter case. Leaving aside the trivial cases $n=1$ and $n=2$, let $n\ge 3$. To rule out $a_n\ne 2,n+1$, assume, if possible, this is the case, and notice that $$a_1-a_n,0,a_1,a_1+a_2,\dots ,a_1+a_2+\cdots +a_{n-1}$$ are pairwise distinct integers, so at least two are congruent modulo $n$. Since $a_n\ne 2,n,n+1$, the first two cannot be congruent modulo $n$, and $(2^{\prime })$ rules out the remaining cases.