= Solution
At any <vertex>, each <complete bipartite graph> $K_{k,k}$ containing it contributes exactly $k$ incident <edges>. Since the copies partition all incident <edges> of the <complete graph>, $n-1$ is a multiple of $k$. In particular $\gcd(n,k)=1$.
If there are $b$ copies, counting their <edges> gives
$$
\frac{n(n-1)}2=bk^2,\qquad n(n-1)=2bk^2.
$$
Thus $k^2\mid n(n-1)$. Since $n$ is coprime to $k^2$, it can be cancelled in this divisibility, yielding
$$
\boxed{k^2\mid n-1.}
$$
If $n=1$, the empty partition is consistent with the same conclusion, since every positive <integer> divides zero.
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