The seesaw is filled with nine pairs of identical twins. All the kids weigh the same (e.g., they all weigh 100, but the actual numbers are irrelevant), except for one set of twins. They are still identical twins, so they weigh the same as each other, but not the same as the other 16 kids (e.g., these two children each weigh 101, but, again, the actual numbers are irrelevant). If (exactly) four kids on one side switch positions with four other kids on the other side to balance the seesaw, which is the odd pair?
The "odd" pair is
the O twins.
The way to balance the seesaw
is to changettl+eoevnn . wweove+leeto
tWO+eLevEn = TweLve+ONe(using capital letters to show the four and four that switched). This cleverly "balances" the seesaw because it spells
TWO+ELEVEN = TWELVE+ONE, i.e., 2+11=12+1. After this switch, the O twins are both third from the end, so the seesaw balances even though they don't weigh the same as the others. That is the only letter that is moved into a symmetric configuration.