National Olympiad Final Round

Let the arithmetic sequence have the first term $a$ and the common difference $d$. The sum of all terms equals the sum of numbers $1,2,\dots ,2017$, i.e., $$\frac{2a+(k-1)d}{2}\cdot k=\frac{2017\cdot 2018}{2}$$ whence $$(2a+(k-1)d)\cdot k=2017\cdot 2018=2\cdot 1009\cdot 2017.$$ Thus the product $2\cdot 1009\cdot 2017$ is divisible by $k$. Since $2$, $1009$, and $2017$ are primes and $k\le 2017$ by assumption, the only possibilities are $k=1$, $k=2$, $k=1009$, and $k=2017$.

All these can occur indeed. A partition into $1$ group trivially satisfies the conditions. An arbitrary partition into $2$ groups also provides two consecutive terms of some arithmetic sequence. Coupling each even number with the next odd number, we get $1008$ groups of size $2$, whose sums are consecutive terms of the arithmetic sequence $5,9,13,\dots$. Forming one additional group containing $1$ as the only element, we obtain a partition satisfying the conditions. Finally, the conditions will also be met by the partition where every integer $1,2,\dots ,2017$ belongs to a separate group.