Balkan Mathematical Olympiad

Induct on $n$ to show that $f(n)=n^3$ for all positive integers $n$. It is readily checked that this $f$ satisfies the conditions in the statement. The base case, $n=1$, is clear. Let $n\ge 2$ and assume that $f(m)=m^3$ for all positive integers $m<n$. Then ${\sum }_{k=1}^{n-1}f(k)=\frac{n^2(n-1{)}^2}{4}$, and reference to the first condition in the statement yields $$f(n)=\sum _{k=1}^{n}f(k)-\sum _{k=1}^{n-1}f(k)={\left(\frac{n(n-1)}{2}+k\right)}^2-\frac{n^2(n-1{)}^2}{4}=k(n^2-n+k),$$ for some positive integer $k$. The divisibility condition in the statement implies $k(n^2-n+k)\le n^3$, which is equivalent to $(n-k)(n^2+k)\ge 0$, showing that $k\le n$. On the other hand, $n^2-n+k$ must also divide $n^3$. But, if $k<n$, then $$n<\frac{n^3}{n^2-1}\le \frac{n^3}{n^2-n+k}\le \frac{n^3}{n^2-n+1}<\frac{n^3+1}{n^2-n+1}=n+1,$$ therefore $\frac{n^3}{n^2-n+k}$ cannot be an integer. Consequently, $k=n$, so $f(n)=n^3$. This completes induction and concludes the proof.