We prove this by showing the given sum, denoted as $S$, is a root of a monic polynomial but not an integer. Then by the rational root theorem, $S$ is irrational.

First we note that $n <\sqrt{n^2+1} < n + \frac{1}{n}$. Hence, the given sum equals $$S = (1001 + k_{1001}) + (1002+k_{1002})+\cdots + (2000+k_{2000})$$

where $0 < k_i < \frac{1}{1001}$. It follows that $$0 < k_{1001} + k_{1002}+\cdots + k_{2000} < 1$$

Hence, $S$ is not an integer.

Next, we are going to show $S$ is a root of a monic polynomial by induction. In fact, we can show that for any positive integers, $a_1$, $a_2$, $\cdots$, $a_n$, the following sum $$S=\sqrt{a_1}+\sqrt{a_2}+\cdots + \sqrt{a_n}$$

is the root of a monic polynomial.

When $n=1$, it is clear that $S_1=\sqrt{a_1}$ is the root of $P_1(x)=x^2 - a_1$.

Assuming $S_k=\sqrt{a_1}+\sqrt{a_k}$ is the root of monic polynomial $P_k(x)$. Then for $n=k+1$, we are going to construct a monic polynomial such that $S_{k+} = S_k +\sqrt{a_{k+1}}$ is one of its roots.

Let $P_k(x) = x^r + c_{r-1}x^{r-1} + c_1x + c_0$. Then $$0=P_k(S_k)=P_k(S_{k+1}-\sqrt{a_{k+1}})=(S_{k+1}-\sqrt{a_{k+1}})^r + c_{r-1}(S_{k+1}-\sqrt{a_{k+1}})^{r-1}+\cdots + c_1(S_{k+1}-\sqrt{a_{k+1}}) + c_0$$

Expanding the right side of the above relation and grouping all the terms of $\sqrt{a_{k+1}}$ leads to $$0=x^r + Q(x) + R(x)\sqrt{a_{k+1}}$$

where the degrees of both $Q(x)$ and $R(x)$ are at most $(r-1)$. Re-arranging the above equation: $$(x^r + Q(x))^2 = a_{k+1}R^2(x)$$

It follows that $S_{k+1}$ is a root of the following equation $$P_{k+1}=(x^r + Q(x))^2 - a_{k+1}R^2(x)$$

It is easy to show that $S_{k+1}$ is monic.