To determine whether the sum $\sqrt{1001^2+1}+\sqrt{1002^2+1}+ \cdots + \sqrt{2000^2+1}$ is a rational number, we assume for the sake of contradiction that it is rational. Since $\sqrt{k}$ is an algebraic integer for each positive integer $k$ and algebraic integers are closed under addition, the given expression must be an algebraic integer. If we assume it is rational, it must be an integer.

Consider the expression:
$$S = (\sqrt{1001^2 + 1} - 1001) + (\sqrt{1002^2+1}-1002) + \cdots + (\sqrt{2000^2+1} - 2000).$$

Each term $\sqrt{k^2+1} - k$ can be approximated as:
$$\sqrt{k^2+1} - k = \frac{1}{\sqrt{k^2+1} + k} < \frac{1}{2k}.$$

Therefore, we have:
$$0 < S < \frac{1}{2} \left( \frac{1}{1001} + \frac{1}{1002} + \cdots + \frac{1}{2000} \right).$$

Using the properties of harmonic series, we can approximate:
$$\frac{1}{1001} + \frac{1}{1002} + \cdots + \frac{1}{2000} \approx \ln\left(\frac{2000}{1000}\right) = \ln 2.$$

Thus:
$$0 < S < \frac{1}{2} \ln 2 < 1.$$

Since $S$ is greater than 0 but less than 1, $S$ cannot be an integer. This contradiction implies that our initial assumption that the sum is rational is false.

Therefore, the sum $\sqrt{1001^2+1}+\sqrt{1002^2+1}+ \cdots + \sqrt{2000^2+1}$ is not a rational number.

The answer is: $\boxed{\text{not a rational number}}$.