Combinatorial proof of $\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$. 
Prove that
  $$\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$$

I can't find counting interpretations for either of the sides. A hint of "if $S$ is a subset of $\{1, . . . , n\}$ and $S^\prime$ is its complement then $|S| + |S^\prime| = n$" was also given, but I still don't know how to begin.
 A: This is not really an answer to the question (very good ones have already been given), but to the more daunting challenge of finding a way to actually use the hint given in the question (in an interesting way).
The left hand side  $L=\sum_{i=0}^n\binom nii$ gives the sum of the sizes of all subsets of an $n$-element set, grouping together the $\binom ni$ subsets of size$~i$. Note that I've thrown in the empty set, which makes no difference for this sum, but makes the number of subsets summed over equal to $2^n$. Since taking the complement of all subsets again gives every subset exactly once (the operation is an involution), $L$ also gives the sum of the sizes of the complements of all subsets of an $n$-element set. But if a set has size $i$, then its complement has size $n-i$ (this is where the hint is used!) so this means that $L=\sum_{i=0}^n\binom ni(n-i)$. 
Adding up the two summations gives 
$$
 2L=\sum_{i=0}^n\binom ni(n+(n-i))=n\sum_{i=0}^n\binom ni=n2^n.
$$
Dividing both sides by$~2$ gives the desired equation $\sum_{i=0}^n\binom nii=L=n2^{n-1}$.
A: We can interpret this combinatorially as the number of ways to form a committee (of any size) with one chairman out of a group of $n$ people.
From $n$ people we first pick a committee of size $i$, then choose one the $i$ committee members to be the chairman.  There are ${n \choose i}$ options for the members of the committee, after which there are $i$ options for the chairman.  If we sum over all $i$ from $1$ to $n$, that covers committees of all possible (nonzero) sizes.  So, we have $\sum_{i=1}^n {n \choose i}i$.
On the other hand, we could first pick one person from the $n$ people to be the chairman.  Then for each of the remaining unchosen $n-1$ people, they can be either in or out of the committee.  That's $2^{n-1}$ possibilities.  So, we have $n2^{n-1}$.
A: Hint: consider the the set of all subsets of $\{1,2,\dots,n\}$ (of which there are $2^n$) and try to find the total sum of the sizes of the subsets in two different ways. For example, the possible subsets of $\{1,2\}$ are $\{\},\{1\},\{2\},\{1,2\}$. Then adding up the sizes of each subset gives $0+1+1+2 = 4$.
More explicitly, if we add up the sizes of all possible subsets of $[n]=\{1,2,\dots,n\}$, we can either:
$1)$ Note that there are $\binom{n}{i}$ subsets of size $i$ which gives that the total sum of sizes is 
$$\sum_{i=1}^{n}\binom{n}{i}i$$
$2)$ Observe that each element is in $2^{n-1}$ subsets, and so contributes $2^{n-1}$ to the total sum. Thus the sum equals 
$$n2^{n-1}$$
The value of the sum doesn't change regardless of how we do it, so the expressions must be the same.
A: Consider an ordered $n+1-tuple$ of numbers,assigned to each subset of the set $S=\{1,2,3,...n,n+1\}$, whose $i$th coordinate is  either $1$ or $0$ accordingly as the the $i$th object is chosen or not respectively. Clearly the left hand side of the equation gives the total number of $1$'s on all the possible $n+1-tuple$s. Again for every $k$ belonging to S a total of $2^n$ $1$s are contributed in those $n+1-tuple$s for $k$.Therefore total no. of $1$s contributed for all $n+1$ numbers is $(n+1)2^n$.Therefore $\mathrm{LHS=RHS}$ Proven.
A: Another possible interpretation is through the expected number of heads in $n$ tosses
of a fair coin which is $n/2$.
That is, we shall have
$$
{n \over 2} = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n} \right)} {i\left( \matrix{
  n \cr 
  i \cr}  \right)\left( {{1 \over 2}} \right)^{\,i} \left( {{1 \over 2}} \right)^{\,n - i} }  = {1 \over {2^{\,n} }}\sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n} \right)} {i\left( \matrix{
  n \cr 
  i \cr}  \right)} 
$$
A: Assume :
You have $n$ beads, numbered $1$ to $n$, of same mass $=1$.
1)$\binom{n}{i}$ is the number of distinct sets of $i$ beads, 
where $i=0,1,2,...n.$
Every set of $i$ beads has a mass $i\cdot1$, there are $\binom{n}{i}$ of them, hence their total mass:
$ T:= \sum_{i=0}^{n}\binom{n}{i}i\cdot 1.$
2)Separate  bead $1$ and consider the total number of sets that can be formed with the remaining $n-1$ beads: $2^{n-1}$.
Hence adding bead $1$ to each of the above sets we conclude:
Bead $1$ is present in $2^{n-1}$ sets out of the $2^n$ original sets.
Bead $1$ contributes $2^{n-1} \cdot 1$ to $T$, so do the beads $2,3,...n $.
Hence : $T=n2^{n-1}\cdot 1$.
