For ranks without considering tie corrections, the well–known upper bound is \[ \sum_{i=1}^{n} d_i^2 \;\le\; \frac{n(n^2-1)}{3}, \] (attained when the two rankings are in reverse order). With only one tie in $x$ the attainable values may be half–integers, but the same upper bound still controls the sum. We need the smallest $n$ such that $19.5\le \dfrac{n(n^2-1)}{3}$. Check $n=3$: $\dfrac{3(9-1)}{3}=8<19.5$. For $n=4$: $\dfrac{4(16-1)}{3}=20\ge 19.5$ (feasible due to the half–integer total from the single tie). Hence the minimum $n$ is $4$, and therefore \[ n^3-n=4^3-4=64-4=\boxed{60}. \]