The question in the title is naively stated, so let be make it more precise: Let $\sum_{n\in\alpha}a_n$ be an ordinal-indexed sequence of real numbers such that $a_n>0$ for each $n\in\alpha$, where $\alpha$ is an ordinal number. What is the smallest $\alpha$ which guarantees that $\sum_{n\in\alpha}a_n$ diverges? Since bijections correspond to rearrangements of the sum, and since the $a_n$ are positive, the sum is either absolutely convergent or diverges to $+\infty$ regardless of the order, so it follows that $\alpha$ is a cardinal number. My intuition tells me that $\alpha=\omega_1$, but I can only prove that $\alpha\le\frak{c}^+$, as follows:
Let $\sum_{n\in\frak{c}^+}a_n$ be a sum of positive reals with $\frak{c}^+$ terms, and let $s_\beta=\sum_{n\in \beta}a_n$ be the sequence of partial sums, so that $s_{\beta+1}=s_\beta+a_\beta$. Then if $\beta<\gamma$, $s_\beta<s_\beta+a_\beta=s_{\beta+1}\le s_\gamma$, so in particular, the $\{s_\beta\}$ are all distinct, and $|\{s_\beta\}|=\frak{c}^+$. If $\sum_{n\in\frak{c}^+}a_n=A$ is finite, then every partial sum is less than $A$, so $\{s_\beta\}\subseteq [0,A]\subseteq\mathbb{R}$, so $\frak{c}^+=|\{s_\beta\}|\le|\mathbb{R}|=\frak{c}$, a contradiction. Thus $\sum_{n\in\frak{c}^+}a_n$ is not finite.
As indicated above, $\alpha\ge\omega_1$ is obvious because $\alpha$ is a cardinal, and $\sum_{n\in\omega}2^{-n}=2$ is finite, so $\alpha>\omega$. Can anyone prove that $\alpha\le\frak{c}$ or that $\alpha=\frak{c}^+$? I can't imagine any set of positive numbers indexed by reals whose sum could possibly be finite, so I lean strongly toward $\alpha\le\frak{c}$, but I don't know how to prove it.
Following André's idea, let me add a small further refinement to Mario's answer: Work in $\mathsf{ZF}$. If $X$ is a set, and $a_x>0$ for each $x\in X$, as usual we define the series $\sum_{x\in X}a_x$ as $$\sup\biggl\{\sum_{x\in X'}a_x\mid X'\subset X\mbox{ is finite}\biggr\}.$$ If the series converges, then we can write $X$ as a countable union of finite sets: $X=\bigcup_{n\in\omega}A_n$, where each $A_n$ is finite. Conversely, for any such set $X$ there are $a_x>0$ for $x\in X$ such that $\sum_{x\in X}a_x$ converges.
Note that the requirement on $X$ is in general strictly weaker than being countable. For example, a Russell set is a set $X$ that can be written in the form $X=\bigcup_n X_n$, where each $X_n$ has size two, the $X_n$ are pairwise disjoint, and no infinite subfamily of the $X_n$ admits a choice function, that is, for any infinite $I\subseteq\omega$, we have that $\prod_{n\in I}X_n$ is empty. It is consistent with $\mathsf{ZF}$ that these sets exist. (The name, of course, comes from Russell's remark that given infinitely many pairs of indistinguishable socks, there is no way to pick a sock from each pair. Curiously, Russell's anecdote actually involved boots rather than socks.)
Suppose first that $X=\bigcup_n A_n$. For each $n$, let $m_n=|A_n|$, and define $\displaystyle a_x=\frac1{m_n 2^n}$ for all $x\in A_n$. We have that $\sum_{x\in X}a_x=2$.
Conversely, suppose that $\sum_{x\in X}a_x$ converges. The point is that, for each positive integer $n$, the set $A_n=\left\{x\in X\mid a_x\in\left[\frac1n,\frac1{n-1}\right)\right\}$ is finite (where $1/0$ is interpreted as $+\infty$). This is because if $|A_n|\ge m$, then $$S=\sum_{x\in X}a_x\ge\sum_{x\in A_n}a_x\ge \frac{m}{n+1},$$ so $S$ diverges if $A_n$ is infinite. But for each $x\in X$, since $a_x>0$, then $x\in A_n$ for some (unique) $n\in\omega$, and it follows that $X=\bigcup_n A_n$.
In fact, for each $n\in\omega$, let $B_n=\{a_x\mid x\in A_n\}$. Note that $B_n$ is the image of a finite set, so it is finite. Moreover, since $B_n$ is a set of reals, it comes equipped with a natural enumeration. For each $x\in X$, let $n_x=|\{y\in X\mid a_y=a_x\}|$. Let $C_n=\bigcup_{x\in A_n}\{a_x\}\times n_x$. Note that $C_n$ is a finite set, and is linearly ordered lexicographically, using the natural ordering of $B_n$ and of the number $n_x=\{m\in\omega\mid m<n_x\}$. It follows that $C=\bigcup_n C_n$ is countable (without any appeal to the axiom of choice). Each $c\in C$ has the form $(a_x,m)$ for some $x\in X$ and some $m<n_x$, and we can define $b_c=a_x$. The point of this is that $\sum_{x\in X}a_x=\sum_{c\in C}b_c$, which is to say that, even if there are convergent series of positive reals indexed by uncountable sets $X$, the fact that $X$ is uncountable is superfluous, as the series can be explicitly rewritten as a countable series.
Finally, let me point out that, if we insist that $X$ is well-ordered (as in Mario's question) then it is countable, as the $A_n$ are naturally ordered by the well-order they inherit from $X$, so $X$ is a countable union of explicitly counted sets. (This shows that if $X$ is an ordinal, then $X<\omega_1$. Note that it is consistent with $\mathsf{ZF}$ that $\omega_1$ is a countable union of countable sets.) Since any countable ordinal can be written as a countable union of finite sets (in fact, of singletons), it follows that $\alpha$, as defined in the question, is indeed $\omega_1$.