% where N is the number of files
for i = 1:N
load(['M' num2str(N) '.m'])
end
This will change based on whether or not you know how many of these files exist (if not, look into using the dir command) and how you want them to be stored (if they're all identically named variables but differ in size, I would suggest a structure; if they're the same size, you could just use a higher-dimension matrix).
