uniformly continuous function $f$ such that $\sum 1/f(n)$ is convergent?

Let us assume that such a function $f$ exists. From the uniform continuity we get a constant $M > 0$ such that $$ \tag 1 |f(n+1) - f(n)| < M $$ for all $n \in \Bbb N$. In particular $|f(n)| < |f(1)| + nM$, so that $\sum_{n=1}^\infty \frac{1}{|f(n)|}$ diverges (as you already observed).

If $\sum_{n=1}^\infty \frac{1}{f(n)}$ is convergent then necessarily $|f(n)| \to \infty$ so that $$ \tag 2 |f(n)| > M $$ for $n \ge n_0$.

Combining these inequalities it follows that for $n \ge n_0$, all $f(n)$ have the same sign, so that $\sum_{n=1}^\infty \frac{1}{f(n)}$ is absolutely convergent, in contradiction to the above observation.

Therefore no such function $f$ exists.