Let $k\ge 1$ be a positive integer.

We consider $4k$ chips, $2k$ of which are red and $2k$ of which are blue. A sequence of those $4k$ chips can be transformed into another sequence by a so-called move, consisting of interchanging a number (possibly one) of consecutive red chips with an
equal number of consecutive blue chips.  For example, we can move from $r\underline{bb}br\underline{rr}b$ to $r\underline{rr}br\underline{bb}b$ where $r$ denotes a red chip and $b$ denotes a blue chip.

Determine the smallest number $n$ (as a function of $k$) such that starting from any initial sequence of the $4k$ chips, we need at most $n$ moves to reach the state in which the first $2k$ chips are red.
1. **Initial Claim and Example:**
   - We claim that at most \( k \) moves are needed to transform any initial sequence of \( 4k \) chips (with \( 2k \) red and \( 2k \) blue) into a sequence where the first \( 2k \) chips are red.
   - To show that there exists a situation that requires \( k \) moves, consider the initial sequence consisting of alternating red and blue chips, i.e., \( rbrbr\cdots brbrb \). This sequence has \( 2k \) disjoint substrings of red chips.

2. **Reduction of Substrings:**
   - Each move can decrease the number of disjoint substrings of red chips by at most 2. Therefore, to reduce \( 2k \) disjoint substrings to 1 contiguous block of \( 2k \) red chips, we need at least:
     \[
     \left\lceil \frac{2k - 1}{2} \right\rceil = k \text{ moves}
     \]

3. **Algorithm to Achieve the Goal:**
   - We now describe an algorithm that ensures the first \( 2k \) chips are red in at most \( k \) moves.

4. **Case Analysis:**
   - **Case 1: There are \( k \) red chips in the first half of the sequence.**
     - In this case, we can spend our \( k \) moves moving chips one at a time from the second half to the first half. This ensures that the first \( 2k \) chips become red.
   - **Case 2: There is a half with more than \( k \) red chips.**
     - If one half has more than \( k \) red chips, we can join all the red chips into one contiguous block by moving chips one at a time from the half with fewer chips to the half with more chips. This takes at most \( k-1 \) moves.
     - If the red chips end up as the last \( 2k \) chips, we can swap the two halves, costing one additional move. Thus, the total number of moves is at most \( k \).

5. **Conclusion:**
   - In both cases, we see that the number of moves required to make the first \( 2k \) chips red is at most \( k \).

\(\blacksquare\)

The final answer is \( \boxed{ k } \).