A finite collection of positive real numbers (not necessarily distinct) is [i]balanced [/i]if each number is less than the sum of the others. Find all $m \ge 3$ 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.
1. **Claim**: All \( m \neq 4 \) work. Clearly, for \( m = 4 \), we have the counterexample \((1, 1, 1, 1)\). Each number is less than the sum of the others, but it cannot be split into three parts with the required property.

2. **Assumption**: Suppose we are given a balanced collection of \( m \neq 4 \) integers. Without loss of generality, assume the sum of all elements is \( 2 \). This is a standard technique to simplify the problem without loss of generality.

3. **Condition**: Given that all numbers in the collection are less than \( 1 \), we need to find a partition into three subcollections, each with a sum less than \( 1 \).

4. **Initial Partition**: Let \( A \) be one of the subcollections with a sum less than \( 1 \) and of maximal sum \( 1 - \delta \). By maximality, all other numbers in the collection are at least \( \delta \).

5. **Case Analysis**:
    - If one of the remaining numbers is greater than \( \delta \), we can choose \( B = \{\delta\} \) and \( C \) as the rest. This satisfies the condition since \( \delta < 1 \) and the sum of \( B \) and \( C \) will be less than \( 1 \).
    - If all other numbers are equal to \( \delta \), then \( \delta = \frac{1}{n} \) for some \( n \) and we have at least \( n + 1 \) numbers equal to \( \delta \).

6. **Recursive Argument**: 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 \( \delta \). Hence, \( m = 2n \) and \( n \geq 3 \).

7. **Final Partition**: For \( m = 2n \) with \( n \geq 3 \), we can choose our three sets to be \( n - 1 \) times, \( n - 1 \) times, and \( 2 \) times \( \delta \). This ensures that each part has a sum less than \( 1 \).

Thus, the only value of \( m \) for which the condition fails is \( m = 4 \).

The final answer is \( \boxed{ m \neq 4 } \).