For those interested, here is the most elementary proof that sets of the natural numbers of positive upper density necessarily have divergent reciprocal sums, r.e. Pete's discussion above.

Suppose $A \subset \mathbf{N}$ and $\displaystyle\limsup_{N \to \infty} \frac{|A\cap [1,N]|}{N} = \alpha > 0$.

Let $N_0=1$, and choose a sequence $N_k$ such that $N_k \geq \frac{4N_{k-1}}{\alpha}$ and $|A \cap [1,N_k)| \geq \frac{\alpha}{2}N_k \text{ }\forall k \in \\mathbf{N}$. $$\sum_{n \in A}\frac{1}{n}= \sum_{k=1}^{\infty}\sum_{n \in A\cap [N_{k-1},N_k)}\frac{1}{n} \geq \sum_{k=1}^{\infty}(|A\cap [1,N_k)|-N_{k-1})\frac{1}{N_k}$$ $$ \geq \sum_{k=1}^{\infty}(\frac{\alpha}{2}N_k - \frac{\alpha}{4}N_k) \frac{1}{N_k}=\sum_{k=1}^{\infty}\frac{\alpha}{4} \to \infty.$$ This is really just a generalization of the classical proof of the divergence of the harmonic series, where you group together progressively larger collections of consecutive terms that add to at least one-half. Hope this helps!

For those interested, here is the most elementary proof that sets of the natural numbers of positive upper density necessarily have divergent reciprocal sums, r.e. Pete's discussion above.

Suppose $A \subset \mathbf{N}$ and $\displaystyle\limsup_{N \to \infty} \frac{|A\cap [1,N]|}{N} = \alpha > 0$.

Let $N_0=1$, and choose a sequence $N_k$ such that $N_k \geq \frac{4N_{k-1}}{\alpha}$ and $|A \cap [1,N_k)| \geq \frac{\alpha}{2}N_k$ for all $k \in \mathbf{N}$. $$\sum_{n \in A}\frac{1}{n}= \sum_{k=1}^{\infty}\sum_{n \in A\cap [N_{k-1},N_k)}\frac{1}{n} \geq \sum_{k=1}^{\infty}(|A\cap [1,N_k)|-N_{k-1})\frac{1}{N_k}$$ $$ \geq \sum_{k=1}^{\infty}(\frac{\alpha}{2}N_k - \frac{\alpha}{4}N_k) \frac{1}{N_k}=\sum_{k=1}^{\infty}\frac{\alpha}{4} \to \infty.$$ This is really just a generalization of the classical proof of the divergence of the harmonic series, where you group together progressively larger collections of consecutive terms that add to at least one-half. Hope this helps!