The answer is $352$ which comes since in the worse case all the balls with the same colour are adjacent to each other and notice that here you have to choose all the balls from $18$ adjacent same-colour groups and $2$ balls from different colour groups to make a totall of $20$ different groups. So in totall the smallest $n$ is $25*18+2=352$

Com10atorics wrote: worse case all the balls with the same colour are adjacent to each other How you prove that this is the worse colouring? As above we prove that $n>=452$ Now colouring the circle and take the smallest number of consecutive balls that contain $20$ different colours. Obviously the first balls colour doesn't appear again otherwise we can put away the first ball but this contradict the minimum of the subject The same for the last ball. So we have to different colours at two corners and $18$inside so the max number of balls is $18*25+2=452$ I am not sure but i have see the same Problem in Turkey at junior.