procedure to pair all people with all others in a group I am looking for an easy method to ensure that all people in a group get to meet all others. The "speed dating" method is to have two rows of people facing each other, and then rotate one of the rows. This works for half of the pairings. How do I get the remaining pairings to happen properly in a simple way?

When $n$ is odd, label each person with elements of $\mathbb Z_{n}$ and, each round of introductions is also labeled from elements in $\mathbb Z_n$. Then in round $k$, person $a$ is introduced to person $b$, if $a+b=k$, with the unique person $i$ so that $2i=k$ not introduced to anybody. This yields $n$ rounds of $\frac{n-1}{2}$ pairs, the best you can do. 
So persons $a$ and $b$ meet in round $a+b$, and person $a$ is left out of the introductions in round $2a$.
When $n$ is even, pick a single individual $X$ out of the set. Then apply the odd case to the $n-1$ other people, but introduce $X$ to the person left out each round. So person $X$ meets person $a$ in round $2a\pmod{n-1}$. This yields $\frac n2$ introductions in $n-1$ rounds, again the best you can do.
So, when $n=6$ we label the people $\{X,0,1,2,3,4\}$, and we get:
Round 0: X0 14 23
Round 1: X3 01 24 
Round 2: X1 02 34
Round 3: X4 03 12
Round 4: X2 04 13

If you absolutely must start with the speed dating approach, with $n_1$ men and $n_2$ women, then after those first $\max(n_1,n_2)$ rounds of speed dating introductions, the rest of the rounds might as well just follow my approach for introducing the $n_1$ men to each other and the $n_2$ women to each other. You can't do better. 
Starting with speed dating would then require $$\max\left(n_1+2\left\lfloor\frac{n_1-1}{2}\right\rfloor+1,n_2+2\left\lfloor\frac{n_2-1}{2}\right\rfloor+1\right)$$
rounds of introductions.
There is a subtle reason that it is easier when $n$ is odd rather than when $n$ is even.
When $n$ is even, we need a binary operation $\star$ on $\{1,2,3,\dots,n\}$ which maps $(i,j)$ to the round number where $i,j$ meet, or $n$ when $i=j$. We obviously need $a\star b=b\star a$ for all $a,b$, and $a\star b=a\star c$ implies $b=c$. There is no good simple arithmetic commutative binary operation on $\mathbb Z_n$ such that $a\star a$ is independent of $a$. (If $n=2^k$, however, we can use the vector space of dimension $k$ over $\mathbb Z_2$ and define $a\star b=a+b$, since $a+a=0$ for all $a\in \mathbb Z_2^k$.)
On the other hand, if $n$ is odd, we have a binary commutative operation on $\mathbb Z_n$ in which $a\star b$ is the round when $a$ and $b$ are introduced, and $a\star a$ is the round when $a$ is left out. If we label the rounds so that $a$ is left out in round $a$, then this means:
$$a\star a = a\\a\star b=b\star a\\a\star b=a\star c\implies b=c$$
Then in any ring in which $2=1+1$ is a unit, this is easily defined as $a\star b=\frac{a+b}{2}$.
