Let the number of blue marbles initially be B: then the number of red marbles initially is 2B. Jill removed n% of the red marbles, that is she removed 2B.n/100, leaving 2B(100-n)/100 reds; she added n% of the blues, that is, she added B.n/100, giving B(100+n)/100 blues. Now we're told that the new number of blues is greater than the new number of reds, that is, B(100+n)/100 > 2B(100-n)/100. So 100+n > 2(100-n), or 3n>100. The smallest integer value of n which satisfies this condition is 34.