Bump......

We use induction on the number of couples to show that $C=1$. Clearly $C>0$. Now we use a permutation of $\{1,1,2,2,\cdots,n,n\}$ to represent the line of dancers. For $n=2$, assuming wlog that at the beginning the leftmost person is 1, there are few possibilities: ${1122},{1212},{1221}$. Whatever the initial order is, we have $C\leq 1$. Now let us suppose we have proved the claim for $n-1$, and consider the first person from the left who is not dancing next to his partner. Call it $A$, and call his partner $B$. If there is an odd number of people on the right of $B$, then at the first step we let $A$ exchange place with the person directly to the right of $B$. Otherwise, since in total there are $2n$ people, there must be an odd number of people on the left of $B$, and we let $A$ exchange place with the person directly to the left of $B$. The inductive hypothesis guarantees that the remaining $n-1$ couples are able to reach the desired configuration in the first step, and from the reasoning above we are also sure that no couple will be divided by $A$ and $B$, hence we're done.

Uhm, I am pretty sure $C=1$ is impossibe for $100$ couples. I think answer is $C=n-1$ where $n$ is the number of couples ( generalized). One may construct an algorithm total of $n-1$ steps. We will prove that $C=n-1$ is minimum by induction. Claim 1.. $C \le n-1$. We will use induction. The cases $n=3,2,1$ are obvious. Now, suppose the claim is true for $n$ and consider total of $n+1$ couples. Our target is to prove $C_{n+1} \le n$. Label peope such that $a_1,a_2,a_3,...,a_{2n}$ where the index $i$ shows $i-th$ person. Write $a_i$ ~ $a_j$ if $a_i,a_j$ are couples. If $a_1$ ~ $a_2$, Then we can just ignore this couple, leaving with $n-1$ couple in random row and hence we are done. Otherwise, Let $a_1$ ~ $a_i$. If $a_i$ is even, swap $a_1$, $a_{i-1}$. Otherwise, swap $a_1,a_{I+1}$. In either case, we will have random row of total $2n-2=2(n-1)$ people and by induction we know that these people can be ordered at maximum $n-1$ swaps, since we already used $1$ swap, $(n-1)+1=n$ therefore we have proved our claim. $\square$ Claim 2. $C_{n} = n-1$ In order to prove the claim, we have to find a order such that we must use at least $n-1$ swaps. Look at the order: $$ b_1,a_2,b_2,a_3,b_3,...,a_n,b_n,1$$Where $a_{i}$ ~ $b_i$. Which is trivial since a swap does not affect other persons, meaning $1$ swap will remove $1$ couple from row all the time, giving total of $n-1$ swaps. Since $n-1 \le C \le n-1$, we have $C=n-1$ therefore we are done $\blacksquare$