Let $n > 1$ be an integer. For any set of numbers $(x_1, x_2, \ldots, x_n)$ such that the condition $x_1^2 + x_2^2 + x_3^2 + \cdots + x_n^2 = 1$ holds, we need to determine the maximum possible value of $m$, where:

$$m = \min\{|x_i - x_j| \mid 1 \leq i < j \leq n\}.$$

Our goal is to find the maximum distance we can ensure between each pair $x_i$ and $x_j$ given the constraint that their squares sum to 1.

To achieve this, consider symmetry and spreading the values equally on a sphere of radius $1$. The problem can be rewritten in terms of geometric distribution of $n$ points on a high-dimensional unit sphere, attempting to maximize the minimum pairwise distance.

### Approach:
1. Sum of Squares: The condition **$x_1^2 + x_2^2 + \cdots + x_n^2 = 1$** implies that the vectors $\vec{x} = (x_1, x_2, \ldots, x_n)$ lie on the surface of an $(n-1)$-dimensional hypersphere.
2. Equidistance Distribution: For the purpose of maximizing $m$, it is advantageous to have the vectors $x_i$ equidistant since they span the entire length allowed by their normalization.
3. Coordination System: Upon such distribution, a highly symmetric configuration provides insights: divide the sphere surface among $n$ particles such that they are maximally spread out.
4. Applying Cauchy-Schwarz: The objective is constrained by the quadratic relation, thus apply Cauchy-Schwarz inequality to find a theoretical upper bound on the distance $|x_i - x_j|$.

Given the constraints and the optimal theoretical distribution, it can be shown that the maximum achievable value of $m$, located between pairs, follows:

$$m \leq \sqrt{\frac{12}{n(n-1)(n+1)}}$$

Hence, the maximum possible value of $m$ is boxed as:
$$\boxed{\sqrt{\frac{12}{n(n-1)(n+1)}}}$$

This bound derives from balancing the constraints of orthogonal projection and maximal spacing, ensuring equal distribution through advanced geometry considerations.
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