Estonia — Open Contests

Answer: The numbers $51, \dots, 100$.

If Juku writes a minus in front of the number $51, \dots, 100$ and a plus in front of the others, then the conditions of the problem are satisfied: for $51 \le n \le 100$, the numbers $n$ and $2n$ have different signs; for $n \le 50$ there is at least one multiple of $n$ among the numbers $51, \dots, 100$.

To show that this arrangement of the signs gives the maximal value of the expression,

consider an arbitrary arrangement of signs satisfying the conditions of the problem. Then always when $100 \ge n \ge 67$ and $n$ has a plus sign, $2n$ must have a minus sign. Also when $66 \ge n \ge 51$ and $n$ has a plus sign, then either $2n$ or $3n$ must have a minus sign. If we change all the pluses in front of the numbers $n$ with $100 \ge n \ge 51$ to minuses, and the minuses in front of the corresponding $2n$ or $3n$ to pluses, then changing minus to plus in front of $m$ corresponds to changing plus to minus in front of $\frac{m}{2}$ or $\frac{m}{3}$ or both. Since $\frac{m}{2} + \frac{m}{3} < m$, the changes increase the value of the expression. Then we can also change all remaining minuses in front of the numbers $1, \dots, 50$ and $101, \dots, 200$ to pluses, which also can only increase the value of the expression. This results in the arrangement of the signs described in the beginning. Hence this arrangement gives the maximal value of the expression.