Answer to old chestnut: four bugs
Each bug walks one foot.
This initially looks like a nasty calculus problem, but there's a simple argument that eliminates any need for calculus.
Initially, each bug has to walk one foot to reach the bug in front of it. Each bug adjusts its direction so it is always facing the bug in front of it. Because of this, each bug is always walking in a direction perpendicular to the one ahead of it and the one behind it. The component of the motion of the bug in front of a given bug in the direction toward/away from it is therefore zero at all times, so the total path length for one bug to reach another remains one foot throughout the walk.
The bugs all meet in the center, by symmetry.