Find the smallest positive integer $n$ for which the polynomial \[x^n-x^{n-1}-x^{n-2}-\cdots -x-1\] has a real root greater than $1.999$.

[i]Proposed by James Lin
1. **Rewrite the polynomial using the geometric series sum formula:**
   The given polynomial is:
   \[
   P(x) = x^n - x^{n-1} - x^{n-2} - \cdots - x - 1
   \]
   Notice that the sum of the geometric series \(1 + x + x^2 + \cdots + x^{n-1}\) is:
   \[
   \frac{x^n - 1}{x - 1}
   \]
   Therefore, we can rewrite the polynomial as:
   \[
   P(x) = x^n - \left(\frac{x^n - 1}{x - 1}\right)
   \]

2. **Simplify the expression:**
   Simplify the expression by combining the terms:
   \[
   P(x) = x^n - \frac{x^n - 1}{x - 1}
   \]
   To combine these terms, we need a common denominator:
   \[
   P(x) = \frac{x^n(x - 1) - (x^n - 1)}{x - 1}
   \]
   Simplify the numerator:
   \[
   P(x) = \frac{x^{n+1} - x^n - x^n + 1}{x - 1} = \frac{x^{n+1} - 2x^n + 1}{x - 1}
   \]

3. **Analyze the simplified polynomial:**
   We need to find the smallest positive integer \(n\) such that \(P(x)\) has a real root greater than \(1.999\). This means we need to solve:
   \[
   x^{n+1} - 2x^n + 1 = 0
   \]
   for \(x > 1.999\).

4. **Substitute \(x = 1.999\) and approximate:**
   Substitute \(x = 1.999\) into the equation:
   \[
   (1.999)^{n+1} - 2(1.999)^n + 1 = 0
   \]
   We can approximate \(1.999\) as \(2\) for simplicity in initial calculations:
   \[
   2^{n+1} - 2 \cdot 2^n + 1 = 0
   \]
   Simplify the equation:
   \[
   2^{n+1} - 2^{n+1} + 1 = 0
   \]
   This simplifies to:
   \[
   1 = 0
   \]
   This approximation is too rough, so we need to be more precise.

5. **Use a more precise approximation:**
   Consider the more precise value \(1.999\):
   \[
   (1.999)^{n+1} - 2(1.999)^n + 1 = 0
   \]
   We need to find the smallest \(n\) such that this equation holds. We can use numerical methods or trial and error to find the smallest \(n\).

6. **Trial and error:**
   Let's try \(n = 10\):
   \[
   (1.999)^{11} - 2(1.999)^{10} + 1 \approx 0
   \]
   Calculate the values:
   \[
   (1.999)^{11} \approx 1.999^{10} \cdot 1.999 \approx 1024 \cdot 1.999 \approx 2047.976
   \]
   \[
   2(1.999)^{10} \approx 2 \cdot 1024 \approx 2048
   \]
   \[
   2047.976 - 2048 + 1 \approx 0
   \]
   This approximation shows that \(n = 10\) is a reasonable solution.

The final answer is \(\boxed{10}\).