Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}<\frac43$

$$\log{k+1\over k}=\int_k^{k+1}{dx\over x}<{1\over k}<\int_{k-1}^k{dx\over x}=\log{k\over k-1}\qquad(k>1)\ .$$ Summing over $k$ produces telescoping series on the left hand and the right hand side. Doing the computations one finds $$\log{3002\over1001}<S<\log{3001\over 1000}\ .$$ Here the left hand side is obviously $>1$. For the right hand side we note that $$e^{4\over3}>1+{4\over3}+{(4/3)^2\over2}={29\over9}>{3001\over1000}\ ,$$ and this shows that $$\log{3001\over 1000}<{4\over3}\ .$$ Update: The OP has desired a calculus-free approach. For the upper estimate one could argue as follows: Using the splitting $$S=\sum_{k=1001}^{1350}{1\over k}+\sum_{k=1351}^{1800}{1\over k}+\sum_{k=1801}^{2400}{1\over k}+\sum_{k=2401}^{3000}{1\over k}+{1\over3001}$$ one obtains $$S<{350\over1000}+{450\over1350}+{600\over1800}+{600\over2400}+{1\over3001}={76\over60}+{1\over3001}<{4\over3}\ .$$

Let $k$ be a fixed integer greater than $0$.

It holds that $\frac{1}{1000+k}\leq \frac{1}{1000+x}$for any $x \in [k-1,k]$.

Integrating for fixed $k \in [1,2001]$ yields $$\frac{1}{1000+k}\leq \int_{k-1}^{k}\frac{1}{1000+x}dx$$ Hence $$\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\ldots+\frac{1}{3001} \leq \int_{0}^{2001}\frac{1}{1000+x}dx=\ln\left(\frac{3001}{1000}\right)$$

And $$\ln\left(\frac{3001}{1000}\right)\sim 1.1 < \frac{4}{3}$$

For a solution without using calculus:

For the upper bound, note for integers $1 \le k \le 1000$:

$$ \frac1{1000+k} + \frac1{3002-k} = \frac{4002}{(1000+k)(3002-k)}= \frac{4002}{4004001-(1001-k)^2} \le \frac{4002}{3004001}$$ where the equality is only when $k=1$.

$$\implies S = \sum_{k=1}^{1000} \left(\frac1{1000+k} + \frac1{3002-k} \right)+\frac1{2001} < \frac{4002}{3004001}\cdot 1000+\frac1{2001} = \frac{8011006001}{6011006001} < \frac43$$

For the lower bound here is another way, using $AM > HM$ for distinct numbers, we get $$\frac1{1000+k} + \frac1{3002-k}> \frac2{2001} \qquad \text{for }k = 1, 2, 3, \dots 1000$$

$$\implies S = \sum_{k=1}^{1000} \left(\frac1{1000+k} + \frac1{3002-k} \right)+\frac1{2001} > \frac{2}{2001} \cdot 1000+\frac1{2001}=1$$