Solution:

Let a run refer to a maximal group of consecutive dinner party guests all of whom are the same type (host or guest). Suppose that there are exactly $k$ runs of hosts and $k$ runs of guests. Let $G_{i}$ and $H_{i}$ denote the number of runs of guests and hosts, respectively, of length exactly $i$. Furthermore, let $X$ denote the number of hosts surrounded by two runs of guests, both of length exactly $1$. We claim that the number of pairs of guests who chat is at least
$$2N - 3k + G_{1} + 2H_{1} + H_{2} - X$$
The number of pairs of guests who chat with no host between them is at least the sum of $\max\{2\ell - 3, 0\}$ over all guest run lengths $\ell$. This sum is at least $2N - 3k + G_{1}$. The number of pairs of guests who chat with exactly two hosts between them is $H_{2}$. Furthermore, the number of pairs of guests who chat with exactly one host between them is at least $2H_{1} - X$. This is because any host surrounded by two runs of guests causes at least two pairs of guests to chat unless these runs are both of length exactly $1$. This proves the claim. Now note that
$$2H_{1} + H_{2} + N \geq 3k$$
because each run of hosts contributes at least three to the left hand side. Furthermore, pairing each run counted in $X$ with the guest run of length $1$ immediately following it in clockwise order shows that $G_{1} \geq X$. Combining these inequalities yields that $2N - 3k + G_{1} + 2H_{1} + H_{2} - X \geq N$, completing the proof of the desired result.