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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 \(N>2\).