Solution:
Let $p(x)$ be a polynomial with the maximum number of minus signs.

$p(x)$ cannot have more than $1005$ minus signs, otherwise $p(1) < 0$ and $p(2) \geq 2^{2010} - 2^{2009} - \ldots - 2 - 1 = 1$, which implies, by the Intermediate Value Theorem, that $p$ must have a root greater than $1$.

Let $p(x) = \frac{x^{2011} + 1}{x + 1} = x^{2010} - x^{2009} + x^{2008} - \ldots - x + 1$. $-1$ is the only real root of $x^{2011} + 1 = 0$ but $p(-1) = 2011$; therefore $p$ has no real roots. Since $p$ has $1005$ minus signs, it is the desired polynomial.

Solution:
Answer: $1005$

Let $p(x)$ be a polynomial with the maximum number of minus signs. $p(x)$ cannot have more than $1005$ minus signs, otherwise $p(1) < 0$ and $p(2) \geq 2^{2010} - 2^{2009} - \ldots - 2 - 1 = 1$, which implies, by the Intermediate Value Theorem, that $p$ must have a root greater than $1$.

Let $p(x) = \frac{x^{2011} + 1}{x + 1} = x^{2010} - x^{2009} + x^{2008} - \ldots - x + 1$. $-1$ is the only real root of $x^{2011} + 1 = 0$ but $p(-1) = 2011$; therefore $p$ has no real roots. Since $p$ has $1005$ minus signs, it is the desired polynomial.