the answer is $ m \leq [\frac{n}{2}]$ for construction, let it be a cyclic graph, number 1,2,..,n clockwise connect (1+m; 1-m);....(n-m;n-3m) (mod n) then we will have each vertex in a cylce with 2m+1 vertices for m = 1 let the graph be a clique for $m = [\frac{n}{2}]$ when n even let the graph be a cyclic graph for example, n = 15 and m = 6 to show that $ m \leq [\frac{n}{2}]$ assume $ m \geq [\frac{n}{2}] + 1$ let a,b,c,d such that distance(a,b) = distance(c,d) = m distance(a,b) and distance(c,d) can't be disjoint paths (since 2m > n) which means any paths with distance = m will intersect other paths so there will be a intersection, let it be e and the paths after vertex e are same because b,d are the max distance from e and we can show that there is no vertex f such that distance (e,f) = m