We claim that all $m \ne 4$ works. Clearly, for $m=4$ we have the counterexample $(1,1,1,1)$. Now suppose that we are given a balanced collection of $m\ne 4$ integers. W.l.o.g. suppose that the sum of all elements is $2$. Thus, we are given that all numbers in the collection are less than $1$ and we need to find a partition into three subcollections, all with sum less than $1$. To start with, let $A$ be one among the subcollections with sum less than $1$ of maximal sum $1-\delta$. By maximality, all other numbers in the collection are at least $\delta$. If one of them is $>\delta$, we win since we can choose $B=\{\delta\}$ and $C$ the rest. So we may suppose that all other numbers are equal to $\delta$. In particular, $\delta=\frac{1}{n}$ for some $n$ and we have at least $n+1$ numbers $\delta$. Repeating this argument with a collection of $n-1$ numbers $\delta$ in place of $A$, we find that indeed all elements in our collection are equal to $\delta$ whence $m=2n$ and $n \ge 3$. But then we may just choose our three sets to be $n-1$ times, $n-1$ times and $2$ times $\delta$ and win.

Clearly, $m=4$ doesn't work at all. Now, assign numbers from $\textit{ balanced}$ with $a_i$ such that $a_{i-1} \leq a_i.$ Now we've got 2 cases: If elememts' number is $2n+1$ then $ (a_{2n+1}),(a_{2n},\cdots,a_2),(a_{2n-1},\cdots,a_1)$ partition works. If element's number is $2n$ and if $(a_{2n}),(a_{2n-1},\cdots,a_1),(a_{2n-2},\cdots,a_2)$ works then we are done. But if it doesn't that means $a_{2n}+a_{2n-2}+\cdots+a_2=a_{2n-1}+a_{2n-3}+\cdots+a_1$ Which gives $a_{2n}=a_{2n-1}$ $$a_{2n-2}=a_{2n-3}$$$$\vdots$$$$a_2=a_1$$ So for $n>2$ $a_{2n}+a_1< a_{2n-1}+\cdots+a_2$. So, $(a_{2n},a_1),(a_{2n-1},\cdots,a_3),(a_{2n-2},\cdots,a_2)$ works.