A finite collection of positive real numbers (not necessarily distinct) is balanced if each number is less than the sum of the others. Find all $m \geq 3$ such that every balanced finite collection of $m$ numbers can be split into three parts with the property that the sum of the numbers in each part is less than the sum of the numbers in the two other parts.

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Answer: The partition is always possible precisely when $m \neq 4$.

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 $m \geq 5$.

Among all possible partitions $A \sqcup B \sqcup C=\{1, \ldots, m\}$ such that

$$
S_{A} \leq S_{B} \leq S_{C},
$$

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

$$
S_{C}-S_{A} \leq g_{c} \leq \frac{1}{2} S_{C}
$$

The first is by the minimality of $S_{C}-S_{A}$, the second by the minimality of $g_{c}$. These two inequalities together yield

$$
S_{A}+S_{B} \geq 2 S_{A} \geq 2\left(S_{C}-g_{c}\right) \geq S_{C}
$$

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

$$
\{1, \ldots, m\}=\{c\} \sqcup B \sqcup(A \cup\{d\}),
$$

resulting in the same sums. A similar procedure applies to $B$. Consequently, $A$ and $B$ must be singleton sets, whence

$$
m=|A|+|B|+|C|=4 \text {. }
$$