If $n=2^r$ is a power of $2$, then $\mathrm{bc}(K_n)\leq n\blog n$. To see this, assign to each vertex $v$ its own vector $x_v\in\{0,1\}^r$, and consider $r=\blog n$ bipartite cliques $H_1,\ldots,H_r$, where two vertices $u$ and $v$ are adjacent in $H_i$ iff $x_u(i)=0$ and $x_v(i)=1$. Since every two distinct vectors must differ in at least one coordinate, each edge of $K_n$ belongs to at least one of these bipartite cliques. Moreover, each of the cliques has weight $(n/2)+(n/2)=n$, since exactly $2^{r-1}=n/2$ of the vectors in $\{0,1\}^r$ have the same value in the $i$-th coordinate. So, the total weight of this covering is $nr=n\blog n$.
Theorem (Hansel-Krichevski 1965): $\mathrm{bc}(K_n)\geq n\blog n$.Proof: We use a probabilistic argument. Let $A_1\times B_1,\ldots,A_t\times B_t$ be a covering of $K_n$ by bipartite cliques. For a vertex $v$, let $m_v$ be the number of these cliques containing $v$. By the double-counting principle, \[ \sum_{i=1}^t(|A_i|+|B_i|)=\sum_{v=1}^nm_v \] is the weight of the covering. So, it is enough to show that the right-hand sum is at least $n\blog n$. To do this, we throw a fair $0$-$1$ coin for each of the cliques $A_i\times B_i$ and remove all vertices in $A_i$ from the graph if the outcome is $0$; if the outcome is $1$, then we remove $B_i$. Let $X=X_1+\cdots+X_n$, where $X_v$ is the indicator variable for the event ``the vertex $v$ survives.'' Since any two vertices of $K_n$ are joined by an edge, and since this edge is covered by at least one of the cliques, at most one vertex can survive at the end. This implies that $E[X]\leq 1$. On the other hand, each vertex $v$ will survive with probability $2^{-m_v}$: there are $m_v$ steps that are ``dangerous'' for $v$, and in each of these steps the vertex $v$ will survive with probability $1/2$. By the linearity of expectation, \[ \sum_{v=1}^n 2^{-m_v} = \sum_{v=1}^n Prob[\mbox{$v$ survives}]= \sum_{v=1}^nE[X_v]=E[X]\leq 1\,. \] The arithmetic-geometric mean inequality states that the arithmetic mean of numbers $a_1,\ldots,a_n$ is at least their geometric mean: \[ \frac{1}{n}\sum_{v=1}^n a_v \geq \Big(\prod_{v=1}^n a_v\Big)^{1/n}\,. \] When applied with $a_v=2^{-m_v}$, this yields \[ \frac{1}{n}\geq \frac{1}{n}\sum_{v=1}^n 2^{-m_v} \geq \Big(\prod_{v=1}^n2^{-m_v} \Big)^{1/n} = 2^{-\frac{1}{n}\sum_{v=1}^n m_v}\,, \] from which $2^{\frac{1}{n}\sum_{v=1}^n m_v}\geq n$, and hence, also $\sum_{v=1}^n m_v\geq n\log_2 n$ follows. $\Box$