Baltic Way shortlist

Answer: The partition is always possible precisely when $m\ne 4$. For $m=3$ it is trivially possible, and for $m=4$ the four equal numbers $g,g,g,g$ provide a counter-example. Henceforth, we assume $m\ge 5$. Among all possible partitions $A\bigsqcup B\bigsqcup C=\{1,\dots ,m\}$ such that $$S_A\le S_B\le S_C,$$ select one for which the difference $S_C-S_A$ is minimal. If there are several such, select one so as to maximise the number of elements in $C$. We will show that $S_C<S_A+S_B$, which is clearly sufficient. If $C$ consists of a single element, this number is by assumption less than the sum of the remaining ones, hence $S_C<S_A+S_B$ holds true. Suppose now $C$ contains at least two elements, and let $g_c$ be a minimal number indexed by a $c\in C$. We have the inequality $$S_C-S_A\le g_c\le \frac{1}{2}{S}_C.$$ The first is by the minimality of $S_C-S_A$, the second by the minimality of $g_c$. These two inequalities together yield $$S_A+S_B\ge 2S_A\ge 2(S_C-g_c)\ge S_C.$$ If either of these inequalities is strict, we are finished. Hence suppose all inequalities are in fact equalities, so that $$S_A=S_B=\frac{1}{2}{S}_C=g_c.$$ It follows that $C=\{c,d\}$, where $g_d=g_c$. If $A$ contained more than one element, we could increase the number of elements in $C$ by creating instead a partition $$\{1,\dots ,m\}=\{c\}\bigsqcup B\bigsqcup (A\cup \{d\}),$$ resulting in the same sums. A similar procedure applies to $B$. Consequently, $A$ and $B$ must be singleton sets, whence $$m=|A|+|B|+|C|=4.$$