Does there exists a positive decreasing sequence $\{a_i\}$ with $\sum_{i\in\mathbb{N}} a_i$ convergent, such that $\forall I\subset\mathbb{N},\sum_{i\in I}a_i$ is an irrational number?
Such an example would give rise to a closed perfect set containing no rationals. I can only do it for infinite $I$ (for example let $a_i=10^{-p_i}$, where $p_i$ is the $i$th prime.), but the set of infinite sums is not closed.
Let $$a_n=\frac{\sqrt{2}}{10^{n!}}.$$ The sum of any finite (non-zero!) number of the $a_i$ is irrational. The sum of an infinite number of the $a_i$ is transcendental, since $\sum_{i=1}^\infty \frac{1}{10^{n_i!}}$ is a Liouville number.
Another easy-to-describe example of a perfect set of irrationals consists of the set of all $x\in(0,1)$ whose continued fraction has the form $$\cfrac1{a_0+\cfrac1{a_1+\cfrac1{a_2+\cfrac1{\dots}}}},$$ where each $a_i$ is either $1$ or $2$. In fact, the set of irrationals in $(0,1)$ is precisely the set of numbers whose continued fraction is infinite. Many perfect sets can be obtained by varying the idea above. Descriptive set-theorists recognize the example at once: The use of continued fractions naturally identifies the set of irrationals with the set $\mathbb N^{\mathbb N}$ of infinite sequences of naturals (sometimes called Baire space). The Cantor set is naturally identified with $\{1,2\}^{\mathbb N}$, and this set corresponds, via continued fractions, to the given example. See here.
For another example, now that we mentioned the Cantor set $C$, it suffices to show that there is a translate of it consisting only of irrationals. See here for this. Curiously, we do not seem to know any concrete examples of reals $r$ such that $C+r$ only contains irrationals.
Regarding your question, an interesting problem is to see what possible sets can be obtained as the set of sums of subseries of a given series. That is, given a sequence $\{x_n\}_{n=1}^\infty$ of positive numbers converging to zero, we consider the set $\Sigma(\{x_n\})$ of all numbers that are the sum of a (finite, or infinite, or even empty) subsequence of $\{x_n\}$. The question is what sets are $\Sigma(\{x_n\})$ for some such sequence $\{x_n\}$.
This question is solved by Guthrie and Nymann, and a nice self-contained exposition of the result can be found in a recent paper by Zbigniew Nitecki:
The term "Cantorval" is due to Pedro Mendes and Fernando Oliveira. A symmetric Cantorval is, by definition, a non-empty compact set $S\subseteq\mathbb R$ such that
Just as all Cantor sets are homeomorphic, so all Cantorvals are homeomorphic as well.
Nitecki's paper, Subsum sets: Intervals, Cantor sets, and Cantorvals, can be downloaded at the ArXiv. It is a nice paper, and I recommend it. It also addresses the case where the sequence is not decreasing (but still converges to zero), or when not all of its terms are positive. This added generality does not change anything: Either one obtains an unbounded interval, containing $0$ (and may equal $\mathbb R$), or a perfect set whose convex hull is $[a,b]$ and is symmetric about the midpoint of this interval, where $a,b$ are the extended reals that are the sum of the negative and positive parts of the sequence, respectively. In this case, the set is again one of the three possibilities listed above.
Note that the only way we only get irrational numbers in $\Sigma(\{x_n\})\setminus\{0\}$ is if $\Sigma(\{x_n\})$ is a Cantor set.
An easy example that this is possible is described in André's answer. I suspect that the set of (non-empty) subseries of $\sqrt2\sum_{n=0}^\infty\frac1{n!}$ is another example, but even though all infinite subseries of $\sum_{n=0}^\infty\frac1{n!}$ are obviously irrational, to show that they are in fact transcendental seems rather more elaborate than to argue in terms of Liouville numbers, as in his answer.
