Revision 2cd44daa-db22-4b1c-b390-dd2774650932 - Code Golf Stack Exchange

# R, 98 bytes
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Very similar to Alex's solution, but uses a substitution rather than a match to determine consecutive letters. Scan is used to get the input and also to split the substitution result on spaces.

 cat(names(a<-table(scan(,'',t=gsub('([A-z]?)(\1円?)[^A-z]*','\\U\2円 ',scan(,''),T,T))))[a==max(a)])

A couple of tests

 > cat(names(a<-table(scan(,'',t=gsub('([A-z]?)(\1円?)[^A-z]*','\\U\2円 ',scan(,''),T,T))))[a==max(a)])
 1: 11 was a race horse, 22 was one too. 11 won one race and 22 one won too.
 19: 
 Read 18 items
 Read 2 items
 O
 > cat(names(a<-table(scan(,'',t=gsub('([A-z]?)(\1円?)[^A-z]*','\\U\2円 ',scan(,''),T,T))))[a==max(a)])
 1: Sally's friends Jimmy and Bobby rummaged for seashells.
 9: 
 Read 8 items
 Read 5 items
 L M
 > cat(names(a<-table(scan(,'',t=gsub('([A-z]?)(\1円?)[^A-z]*','\\U\2円 ',scan(,''),T,T))))[a==max(a)])
 1: 11ss11nn
 2: 
 Read 1 item
 Read 2 items
 N S
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