Is there a minimal diverging series? Is there a function $f:\mathbb{N} \to \mathbb{R}^+$ s.t. its series $\Sigma_{i=0}^\infty f(n)$ diverges but the series for all function in $o(f)$ converge?

No. Suppose $\sum_{i=0}^\infty f(i)=\infty$. Then we can find values $N_j\in \mathbb N$ such that 
$$\sum_{i=N_j}^{N_{j+1}-1}f(i)>1$$
for all $j\in\mathbb N$. Define $g:\mathbb N\to\mathbb R^+$ by
$$g(i)=\begin{cases}f(i) &\text{if } i<N_1\\
\frac{1}{2}f(i) &\text{if } N_1\leq i<N_{2}\\
\frac{1}{3}f(i) &\text{if } N_2\leq i<N_{3}\\
&\vdots
\end{cases}$$
and note that $g\in o(f)$ yet $\sum_{i=0}^\infty g(i)\geq \sum_{j=1}^\infty \frac{1}{j}=\infty$.
