From what I understood from the original post, this should work (probably a long-handed version). Change *.txt to whatever you need to, to grab each file.

#!/usr/bin/env python
import glob
from collections import defaultdict(list)

def open_file(file):
    with open(file, 'r') as f:
            for line in f:
                ids[line.strip().split('\t')[0]].append(int(line.strip().split('\t')[1]))

ids = defaultdict(list)
nfiles = 0
for file in glob.glob('*.txt'):
    nfiles += 1
    open_file(file)

for i in ids:
    if len(ids[i][0]) == 3:
            if len(set(ids[i][0])) == nfiles:
                print i, '\t', '\t'.join(ids[i][0])
            else:
                    if max[ids[i][0]] - min[ids[i][0]] <= 20:
                        print i, '\t', '\t'.join(ids[i][0])

If you're using Python, you could use a set to build a list of keys common to all inputs, then do a second pass over inputs to write filtered outputs. Something like the following (untested) script:

#!/usr/bin/env python

in_fns = ['in1.txt', 'in2.txt', 'in3.txt']
# open read-only input handles
in_fhs = [open(in_fn, 'r') for in_fn in in_fns]
ks = set()
# populate ks with keys from each file
for in_handle in in_fhs:
    for line in in_handle:
        line = line.strip()
        (k, v) = line.split()
        ks.add(k)
    in_handle.close()
# open fresh read-only input handles
in_fhs = [open(in_fn, 'r') for in_fn in in_fns]
out_fns = ['out1.txt', 'out2.txt', 'out3.txt']
# open write-only output handles
out_fhs = [open(out_fn, 'w') for out_fn in out_fns]
# write filtered output based on contents of ks set
for idx, in_handle in enumerate(in_fhs):
    for line in in_handle:
        line = line.strip()
        (k, v) = line.split()
        if k in ks:
            out_fhs[idx].write("%s\n" % (line))
    in_handle.close()         
    out_fhs[idx].close()

This could be done with a command-line approach, but you asked for Python or Perl, so this is one way to do it.