Find the least positive integer $n$, such that there is a polynomial \[ P(x) = a_{2n}x^{2n}+a_{2n-1}x^{2n-1}+\dots+a_1x+a_0 \] with real coefficients that satisfies both of the following properties:
      - For $i=0,1,\dots,2n$ it is $2014 \leq a_i \leq 2015$.
      - There is a real number $\xi$ with $P(\xi)=0$.
To find the least positive integer \( n \) such that there exists a polynomial 
\[ P(x) = a_{2n}x^{2n} + a_{2n-1}x^{2n-1} + \dots + a_1x + a_0 \]
with real coefficients satisfying:
1. \( 2014 \leq a_i \leq 2015 \) for \( i = 0, 1, \dots, 2n \),
2. There is a real number \( \xi \) such that \( P(\xi) = 0 \),

we proceed as follows:

1. **Constructing a Polynomial for \( n = 2014 \)**:
   - Consider \( n = 2014 \).
   - We can construct a polynomial \( P(x) \) with alternating coefficients \( a_{2k} = 2014 \) and \( a_{2k+1} = 2015 \) for \( k = 0, 1, \dots, 2013 \).
   - Specifically, let:
     \[ P(x) = 2014x^{4028} + 2015x^{4027} + 2014x^{4026} + \dots + 2015x + 2014. \]
   - Evaluate \( P(-1) \):
     \[ P(-1) = 2014(-1)^{4028} + 2015(-1)^{4027} + 2014(-1)^{4026} + \dots + 2015(-1) + 2014. \]
     Since \( (-1)^{2k} = 1 \) and \( (-1)^{2k+1} = -1 \), we get:
     \[ P(-1) = 2014 - 2015 + 2014 - 2015 + \dots + 2014 - 2015 + 2014. \]
     The terms cancel out pairwise, leaving:
     \[ P(-1) = 2014 - 2015 = -1. \]
     Thus, \( P(-1) = 0 \) is not satisfied. We need to reconsider the construction.

2. **Analyzing the Polynomial for \( n < 2014 \)**:
   - Assume there exists a polynomial \( P(x) \) of degree \( 2n \) with \( n < 2014 \) that has a real root.
   - For \( x \ge 0 \), \( P(x) \ge 2014 > 0 \), so the root must be negative.
   - Let \( Q(x) = P(-x) \). Then \( Q(x) \) must have a positive real root.
   - Consider:
     \[ Q(x) = 2014x^{2n} - 2015x^{2n-1} + 2014x^{2n-2} - \dots - 2015x + 2014. \]
   - Define:
     \[ R(x) = 2014(x^{2n} + x^{2n-2} + \dots + 1) - 2015(x^{2n-1} + x^{2n-3} + \dots + x). \]
   - We need to show \( R(x) > 0 \) for \( x > 0 \).

3. **Proving \( R(x) > 0 \) for \( x > 0 \)**:
   - Evaluate \( R(1) \):
     \[ R(1) = 2014(n+1) - 2015n = 2014 - n. \]
     For \( n < 2014 \), \( R(1) > 0 \).
   - For \( x \ne 1 \), consider:
     \[ 2014 \frac{x^{2n+2} - 1}{x^2 - 1} \ge 2015x \frac{x^{2n} - 1}{x^2 - 1}. \]
     Simplifying, we get:
     \[ \frac{x^{2n+2} - 1}{x(x^{2n} - 1)} \ge 1 + \frac{1}{2014}. \]
   - For \( n < 2014 \), it suffices to show:
     \[ \frac{x^{2n+2} - 1}{x(x^{2n} - 1)} \ge \frac{n+1}{n}. \]
   - This is equivalent to:
     \[ nx^{2n} + nx^{2n-2} + \dots + nx^2 + n \ge (n+1)x^{2n-1} + (n+1)x^{2n-3} + \dots + (n+1)x. \]

4. **Using AM-GM Inequality**:
   - For any \( k \), \( nx^{2k+1} \le \frac{n}{2}x^{2k} + \frac{n}{2}x^{2k+2} \) by AM-GM.
   - Additionally, \( x^k + x^{2n-k} \le x^{2n} + 1 \).
   - Summing these inequalities for all \( k \) from \( 0 \) to \( n-1 \) and for all odd \( k \) from \( 1 \) to \( 2n-1 \), we get the desired result.

Thus, for \( n < 2014 \), such a polynomial cannot exist. Therefore, the least positive integer \( n \) is \( \boxed{2014} \).