You are given $N$ such that $ n \ge 3$. We call a set of $N$ points on a plane acceptable if their abscissae are unique, and each of the points is coloured either red or blue. Let's say that a polynomial $P(x)$ divides a set of acceptable points either if there are no red dots above the graph of $P(x)$, and below, there are no blue dots, or if there are no blue dots above the graph of $P(x)$ and there are no red dots below. Keep in mind, dots of both colors can be present on the graph of $P(x)$ itself. For what least value of k is an arbitrary set of $N$ points divisible by a polynomial of degree $k$?
1. **Understanding the Problem:**
   We need to find the least degree \( k \) of a polynomial \( P(x) \) such that any set of \( N \) points on a plane, with unique abscissae and colored either red or blue, can be divided by \( P(x) \). The polynomial \( P(x) \) divides the points if:
   - No red points are above the graph of \( P(x) \) and no blue points are below it, or
   - No blue points are above the graph of \( P(x) \) and no red points are below it.

2. **Construction of Polynomial:**
   Consider a polynomial \( P(x) \) of degree \( \leq N-2 \) passing through \( N-1 \) of the points. By Lagrange Interpolation, such a polynomial exists. This polynomial can be constructed to ensure that it divides the points as required.

3. **Proof of Converse:**
   To prove that \( k = N-2 \) is the least degree, we need to show that for any set of \( N \) points, a polynomial of degree \( \leq N-3 \) cannot always divide the points.

4. **Example Construction:**
   Consider the points \((a_1, y_1), (a_2, y_2), \ldots, (a_N, y_N)\) where \( a_1 < a_2 < \cdots < a_N \). Assume the points are colored such that:
   - Red points are at \((a_1, y_1)\) and \((a_3, y_3)\),
   - Blue points are at \((a_2, y_2)\), \((a_4, y_4)\), \((a_5, y_5)\), etc.

5. **Properties of Polynomial \( P(x) \):**
   - \( P(x) \) must be non-constant.
   - For \( P(x) \) to divide the points, it must satisfy \( P(a_i) \cdot P(a_{i+1}) \leq 0 \) for all \( i = 1, 2, \ldots, N-2 \).

6. **Root Analysis:**
   Each interval \([a_i, a_{i+1}]\) must contain a root of \( P(x) \). Define \( a_i \) as *good* if it is a root of \( P(x) \) and *bad* otherwise. If there are \( \leq 1 \) bad numbers, we are done. Otherwise, let \( S = \{i : a_i \text{ is bad} \} = \{b_1, b_2, \ldots, b_t \} \) with \( t \geq 2 \).

7. **Interval Analysis:**
   The interval \([a_{b_i}, a_{b_{i+1}}]\) must contain at least \( b_{i+1} - b_i \) roots of \( P(x) \) with multiplicity, as it contains \( b_{i+1} - b_i + 1 \) good numbers. Another root must be there because of the inequality condition on \( P(x) \).

8. **Conclusion:**
   This implies \( P(x) \) has at least \( N-2 \) roots. Since \( P(x) \) is non-constant, the degree of \( P(x) \) must be at least \( N-2 \).

\(\blacksquare\)

The final answer is \( \boxed{ N-2 } \)