100 couples are invited to a traditional Modolvan dance. The $200$ people stand in a line, and then in a $\textit{step}$, (not necessarily adjacent) many swap positions. Find the least $C$ such that whatever the initial order, they can arrive at an ordering where everyone is dancing next to their partner in at most $C$ steps.
1. **Base Case:**
   We start by verifying the base cases for small values of \( n \).

   - For \( n = 1 \): There is only one couple, and they are already next to each other. Thus, \( C_1 = 0 \).
   - For \( n = 2 \): There are two couples. If they are not initially next to each other, we can swap them in one step. Thus, \( C_2 = 1 \).
   - For \( n = 3 \): There are three couples. We can arrange them in at most two steps. Thus, \( C_3 = 2 \).

2. **Inductive Hypothesis:**
   Assume that for \( n \) couples, the maximum number of steps required to arrange them such that each person is next to their partner is \( n-1 \). That is, \( C_n \leq n-1 \).

3. **Inductive Step:**
   We need to show that for \( n+1 \) couples, the maximum number of steps required is \( n \). That is, \( C_{n+1} \leq n \).

   - Label the people as \( a_1, a_2, a_3, \ldots, a_{2(n+1)} \), where \( a_i \sim a_j \) indicates that \( a_i \) and \( a_j \) are a couple.
   - If \( a_1 \sim a_2 \), then this couple is already next to each other. We can ignore this couple and focus on the remaining \( n \) couples. By the inductive hypothesis, these \( n \) couples can be arranged in \( n-1 \) steps. Since we did not need any additional steps for the first couple, the total number of steps is \( n-1 \).
   - If \( a_1 \sim a_i \) for some \( i \neq 2 \), we need to bring \( a_i \) next to \( a_1 \). If \( i \) is even, swap \( a_1 \) and \( a_{i-1} \). If \( i \) is odd, swap \( a_1 \) and \( a_{i+1} \). This ensures that \( a_1 \) and \( a_i \) are now adjacent. This requires one step.
   - After this swap, we are left with \( 2n \) people (i.e., \( n \) couples) to arrange. By the inductive hypothesis, these \( n \) couples can be arranged in \( n-1 \) steps. Since we already used one step, the total number of steps is \( (n-1) + 1 = n \).

4. **Conclusion:**
   By induction, we have shown that \( C_{n+1} \leq n \). Therefore, \( C_n = n-1 \) for all \( n \).

5. **Lower Bound:**
   To show that \( C_n = n-1 \) is the minimum, consider the worst-case scenario where the couples are arranged as \( b_1, a_2, b_2, a_3, b_3, \ldots, a_n, b_n \), where \( a_i \sim b_i \). Each swap can only bring one couple together, so we need at least \( n-1 \) swaps to arrange all \( n \) couples.

Since \( n-1 \leq C \leq n-1 \), we conclude that \( C = n-1 \).

\(\blacksquare\)

The final answer is \( \boxed{ C = n-1 } \).