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