Theorem (Sergeev [3]): \OR_2(K)=\Omega(n\log n).Proof: Recall that rows and columns of K are labeled by subsets of [r]=\{1,\ldots,r\}, where n=2^r is the dimension of K. For 0\leq i\leq r, let A_i the m_i\times m_i sumbatrix of K with m_i=\tbinom{r}{i}, whose rows are labeled by subsets of size i, and columns by subsets of size r-i. Note that the submatrices A_i do not share common rows or columns. This implies \OR_2(K)\geq \sum_{i=0}^r\OR_2(A_i).
On the other hand, each of the submatrices A_i has 0s on the diagonal, and 1s elsewhere. That is, each A_i is just the complement \overline{I}_{m_i} of the identity m_i\times m_i matrix I_{m_i}. The matrix \overline{I_m} is the adjacency matrix of the complete (non-bipartite) graph \mathrm{K}_m on m vertices. Well-known results of Hansel and Krichevski (see here) imply that every covering of \mathrm{K}_m by bicliques (bipartite complete subgraphs) must have weight m\log m; the weight of a biclique is the number of its vertices, and the weight of a covering is the sum of weights of its bicliques. When translated to the language of OR-circuits, this implies that \OR_2(A_i)\geq m_i\log m_i for all i=1,\ldots,r-1. Thus, \OR_2(K)\geq \sum_{i=1}^{r-1}m_i\log m_i=\sum_{i=1}^{r-1}\binom{r}{i}\log \binom{r}{i} \geq \sum_{i=1}^{r-1}\binom{r}{i}\log \Big(\frac{r}{i}\Big)^i = \sum_{i=1}^{r-1}\binom{r}{i} i \log \frac{r}{i} =\Omega(r2^r)=\Omega(n\log n). \Box