First, we note that we cannot have $n \leq 6$, since the first 6 letters are X's. After 6 X's and 3 Y's, there are twice as many X's as Y's. In this case, $n=6+3=9$. After 6 X's and 12 Y's, there are twice as many Y's as X's. In this case, $n=6+12=18$. The next letters are all Y's (with 24 Y's in total), so there are no additional values of $n$ with $n \leq 6+24=30$. At this point, there are 6 X's and 24 Y's. After 24 Y's and 12 X's (that is, 6 additional X's), there are twice as many Y's as X's. In this case, $n=24+12=36$. After 24 Y's and 48 X's (that is, 42 additional X's), there are twice as many X's as Y's. In this case, $n=24+48=72$. Since we are told that there are four values of $n$, then we have found them all, and their sum is $9+18+36+72=135$.