Can we add an uncountable number of positive elements, and can this sum be finite? Can we add an uncountable number of positive elements, and can this sum be finite? 
I always have trouble understanding mathematical operations when dealing with an uncountable number of elements. Any help would be great. 
 A: We have the next proposition
Proposition 1. Let $X$ be an at most countable set, and let $f\colon X\to\mathbf R$ be a function. Then the series $\sum_{x\in X} f(x)$ is absolutely convergent if and only if $$\sup\left\{\sum_{x\in A}|f(x)|:A\subseteq X,  A\text{ finite}\right\}<\infty.$$
Inspired  by  this  proposition,  we  may  now  define  the  concept  of an  absolutely  convergent  series  even  when  the  set  $X$  could  be uncountable.
Definition 2. Let $X$ be a set (which could be uncountable), and let $f\colon X\to\mathbf R$ be a function. We say that the series $\sum_{x\in X} f(x)$ is absolutely convergent if and only if $$\sup\left\{\sum_{x\in A}|f(x)|:A\subseteq X,  A\text{ finite}\right\}<\infty.$$
Note  that  we  have  not  yet  said  what  the  series $\sum_{x\in X} f(x)$ is equal  to.  This  shall  be  accomplished  by  the  following proposition.
Proposition 3. Let  $X$  be  a  set  (which  could  be  uncountable),  and let  $f\colon X\to\mathbf R$ be  a  function  such  that  the  series  $\sum_{x\in X} f(x)$  is absolutely  convergent.  Then  the set $\{x\in X:f(x)\ne0\}$  is  at  most countable. 
Because  of  this,  we  can  define  the  value  of $\sum_{x\in X} f(x)$  for  any absolutely  convergent  series  on  an  uncountable  set  $X$  by  the  formula $$\sum_{x\in X} f(x):=\sum_{x\in X:f(x)\ne0} f(x),$$ since  we  have  replaced  a  sum  on  an  uncountable  set  $X$  by  a  sum on  the  countable  set $\{x\in X:f(x)\ne0\}$.  (Note  that  if  the former  sum  is  absolutely  convergent,  then  the  latter  one  is  also.) Note  also  that  this  definition  is  consistent  with  the  definitions  for  series  on  countable  sets.
Remark. The definition of series on countable sets that are use is
Definition 4. Let  $X$  be  a  countable set,  and  let  $f\colon X\to\mathbf R$  be  a  function.  We  say  that  the  series $\sum_{x\in X}f(x)$  is  absolutely  convergent  iff  for  some  bijection  $g\colon\mathbf N\to X$,  the  sum $\sum_{n=0}^\infty  f(g(n))$  is  absolutely  convergent.  We  then  define the  sum  of $\sum_{x\in X}f(x)$ by  the  formula $$\sum_{x\in X}f(x)=\sum_{n=0}^\infty  f(g(n)).$$
A: Suppose $\{s_i : i\in\mathcal I\}$ is a family of positive numbers.$^\dagger$  We can define
$$
\sum_{i\in\mathcal I} s_i = \sup\left\{ \sum_{i\in\mathcal I_0} s_i : \mathcal I_0 \subseteq \mathcal I\ \&\ \mathcal I_0 \text{ is finite.} \right\}
$$
(If both positive and negative numbers are involved, then we have to talk about a limit rather than about a supremum, and then the definition is more complicated and we have questions of conditional convergence and rearrangements.)
Now consider
\begin{align}
& \{i\in\mathcal I : s_i \ge 1\} \\[4pt]
& \{i\in\mathcal I : 1/2 \le s_i < 1 \} \\[4pt]
& \{i\in\mathcal I : 1/3 \le s_i < 1/2 \} \\[4pt]
& \{i\in\mathcal I : 1/4 \le s_i < 1/3 \} \\[4pt]
& \quad \quad \quad \vdots
\end{align}
If one of these sets is infinite, then $\sum_{i\in\mathcal I} s_i=\infty$.  But if all are finite, then $\mathcal I$ is at most countably infinite.
Thus the sum of uncountably many positive numbers is infinite.
I don't know whether by some arguments about rearrangements one could somehow have some sensible definition of a sum of numbers not all having the same sign that could give us a somehow well defined sum of uncountably many numbers and get a finite number.

$^\dagger$ In the initial edition of this answer, I said "Let $S$ be a set of positive numbers and then went on to say
$$
\sum S = \left\{ \sum S_0 : S_0\subseteq S\ \&\ S_0\text{ is finite.} \right\}
$$
However, Dustan Levenstein pointed out in comments that "this definition fails to allow for the same number to occur twice in a sum".  Rather than "twice", I'd say "more than once", since a number might even occur an uncountably infinite number of times.
A: Let $H$ be a positive unlimited integer of nonstandard analysis. Then, for example, the sum
$$ \sum_{n=1}^H n = \frac{H(H+1)}{2}$$
is a sum of uncountably many positive numbers... but it's a hyperfinite nonstandard sum, so it exists by the usual methods of nonstandard analysis. The sum is unlimited, though. Other sums can be finite: e.g.
$$ \sum_{n=1}^H \frac{1}{n!}$$
is a finite nonstandard real number that is infinitesimally close to $e$.
That said, IMO, thinking of hyperfinite sums from nonstandard analysis as being sums of uncountably many elements isn't a particularly fruitful line of thought. (also, the sum only works for internal sequences of elements anyways; you can't take an arbitrary uncountable collection)

I bring this up mainly to show that uncountable sums can make sense in some contexts, even if you can't really do much in a standard setting. Each summation operator one might define can have its own sorts of pecularities.
A: Here is the more general definition of the sum of any function $f:I\rightarrow E$, where $I=$any non empty set and $E$ is a real or complex topological vector space.
Definition 1 We say that the family $f=(f(i))_{i\in I}$ is summable in $E$ of sum $\omega\in E$ if and only if, for all neighborhood $V$ of $\omega$, there exists a finite subset $A\subset I$ of $I$ such that for all finite parts $K$ of $I$ containing $A$, we have $\sum_{i\in L}f(i)-\omega\in V$. f is said to be absolutely summable if and only if $\|f\|=(\|f(i)\|)_{i\in I}$ is summable in $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$.
Remark 2 This definition does not guarantee the uniqueness of the sum of a given family. The topological vector space $E$ mus be separable for the uniqueness to hold, in which case we write
$$\sum_{i\in I}f(i)=\omega.$$
Proposition 3 If $E$ is a normed space, then a family $f=(f(i))_{i\in I}$ is summable of sum $\omega$ if and only if for any $0<\varepsilon\in \mathbb{R}$ there exists a finite part $A$ of $I$ such that for all finite part $K$ of $I$ with $A\subset K$, we have:
$$\|\sum_{i\in K}f(i)-\omega\|<\varepsilon.$$
Remark 4 If $f=(f(i))_{i\in I}$ is a family in $E$, we can definie $\mathcal{F}(I)=\{A\subset I:~A~\text{is finite}\}$. Embedding $\mathcal{F}(I)$ with the direction ``$\leq$'' defined by $A\leq B$ in $\mathcal{F}$ iff $A\subset B$, we make $\mathcal{F}(I)$ a directed set (or Riesz space). The if we definite $S:\mathcal{F}(I)\rightarrow E$, by $$S(K)=\sum_{i\in K} f(i),$$ we get a net in $E$. We have the following:
Proposition 5 $f$ is summable if and only if the set $S(K)$, defined above, is convergent. 
Notice that here we have defined the sum of any kind of number, countable or uncountable. But it is proved that in case $E=\mathbb{F}$ and $I=\mathbb{N}$, the summability coincide with the absolute summability, which in turns coincide with the commutative convergence of the series $\sum_{n=1}^{+\infty}f(n)$, and the value of the sum of the family is the same as the value of this series.
