re = /(\b\d{8})(\d{6}Z)\b/
str = '20150624145707Z
20150603182904Z
20150421163629Z
20140529152951Z
20091015212450Z
20140428144528Z
20081009222929Z
20150817231245Z
20150420184902Z
20140428143953Z
20150315070002Z
20150315070003Z
20150212154444Z
20150527165706Z
19970205203358Z
20150413180400Z
20150924180348Z
20030823025726Z
20150423223718Z
20141022171412Z
20150412232614Z
20070505011754Z
20140428144155Z
20141231221928Z
20150305150408Z
20060103155521Z
20131114224409Z
20150910000720Z
20150831151016Z
'
subst = '$1'
result = str.gsub(re, subst)
# Print the result of the substitution
puts result
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Ruby, please visit: http://ruby-doc.org/core-2.2.0/Regexp.html