Find the least positive integer \(M\) for which there exist a positive integer \(n\) and polynomials \(P_1(x)\), \(P_2(x)\), \(\ldots\), \(P_n(x)\) with integer coefficients satisfying \[Mx=P_1(x)^3+P_2(x)^3+\cdots+P_n(x)^3.\]

[i]Proposed by Karthik Vedula[/i]
1. **Claim 1: \(3 \mid M\)**

   Differentiate the given equation \(Mx = P_1(x)^3 + P_2(x)^3 + \cdots + P_n(x)^3\) with respect to \(x\):
   \[
   M = \sum_{i=1}^n 3P_i'(x)P_i(x)^2.
   \]
   Each term on the right-hand side is \(3\) times an integer polynomial, hence \(3 \mid M\).

2. **Claim 2: \(2 \mid M\)**

   Consider the roots of the polynomial \(x^2 + x + 1 = 0\), which are \(\alpha = -\frac{1}{2} + \frac{i\sqrt{3}}{2}\) and \(\beta = -\frac{1}{2} - \frac{i\sqrt{3}}{2}\). Evaluate the given equation at \(x = \alpha\) and \(x = \beta\):
   \[
   M\alpha = \sum_{i=1}^n P_i(\alpha)^3 \quad \text{and} \quad M\beta = \sum_{i=1}^n P_i(\beta)^3.
   \]
   Adding these two equations, we get:
   \[
   M(\alpha + \beta) = \sum_{i=1}^n \left( P_i(\alpha)^3 + P_i(\beta)^3 \right).
   \]
   Since \(\alpha + \beta = -1\), we have:
   \[
   -M = \sum_{i=1}^n \left( P_i(\alpha)^3 + P_i(\beta)^3 \right).
   \]
   Reduce \(P_i(\alpha)\) to the linear integer polynomial \(r_i\alpha + s_i\) via repeated applications of the identity \(\alpha^2 = -\alpha - 1\). Since \(\alpha\) and \(\beta\) are conjugates, it follows that \(P_i(\beta) = r_i\beta + s_i\) as well. Thus,
   \[
   -M = \sum_{i=1}^n \left( (r_i\alpha + s_i)^3 + (r_i\beta + s_i)^3 \right).
   \]
   Expanding the cubes, we get:
   \[
   (r_i\alpha + s_i)^3 + (r_i\beta + s_i)^3 = 2r_i^3 + 2s_i^3 - 3r_is_i(r_i + s_i).
   \]
   Observe that \(r_is_i(r_i + s_i)\) is always even, so every term in the right-hand side summation is even, hence \(2 \mid M\).

3. **Finding the least \(M\)**

   We need to find the least positive integer \(M\) such that \(M\) is divisible by both 2 and 3. The least common multiple of 2 and 3 is 6. We need to verify if \(M = 6\) works:
   \[
   6x = (x)^3 + (-x)^3 + (x + 1)^3 + (x - 1)^3.
   \]
   Simplifying the right-hand side:
   \[
   6x = x^3 + (-x)^3 + (x + 1)^3 + (x - 1)^3 = x^3 - x^3 + (x + 1)^3 + (x - 1)^3.
   \]
   \[
   (x + 1)^3 = x^3 + 3x^2 + 3x + 1 \quad \text{and} \quad (x - 1)^3 = x^3 - 3x^2 + 3x - 1.
   \]
   Adding these:
   \[
   (x + 1)^3 + (x - 1)^3 = (x^3 + 3x^2 + 3x + 1) + (x^3 - 3x^2 + 3x - 1) = 2x^3 + 6x.
   \]
   Therefore,
   \[
   6x = 6x.
   \]
   This confirms that \(M = 6\) works.

The final answer is \(\boxed{6}\).