WLOG let the first bowl have the smaller average. If N>2, then x=12 is obviously achievable - for instance, start with [1],[2,3,…N] and move the 2 ball. It's trivial to check both averages increase by 12.
Now assume x>12 is possible for some N, and let the first bowl initially have k balls in it, summing to S. Obviously S≥1+2+…k=k(k+1)2. Further let the ball moved be of value a.
The average of the first bowl increases by S+ak+1−Sk>12.
The average of the second bowl increases by N(N+1)2−S−aN−k−1−N(N+1)2−SN−k>12.
Solving for S yields S<k(k+1)2, contradiction.
Hence x=12 for all N>2.