We can adapt the proof of Graham-Pollak slightly. If our good sets are $R_k\times C_k$, let us instantiate $x_i, y_i\in \mathbb R$ for $i=1,..,n$ and define $x(S) := \sum_{i\in S} x_i$ and likewise $y(S)$. Then if the good sets ($K$ of them) indeed form a partition, we have $$\sum_{i=1}^K x(R_i) y(C_i) = \sum_{i\neq j} x_iy_j = (\sum x_i)(\sum y_i) - \sum x_i y_i$$so if $K\le n-1$ we can find a nontrivial solution to the system $X(R_i) = 0$, but because $(\sum x_i)(\sum y_i) - \sum x_iy_i$ for any choice of $y$, this means that $x_1=x_2=...=x_n = \sum x_i =0$, a contradiction. The construction for $K=n$ is easy.