Number of vectors so that no two subset sums are equal Consider all $10$-tuple vectors each element of which is either $1$ or $0$. It is very easy to select a set $v_1,\dots,v_{10}= S$ of $10$ such vectors so that no two distinct subsets of vectors $S_1 \subset S$ and $S_2 \subset S$  have the same sum. Here $\sum_{v \in S_i} v$ assumes simple element-wise vector addition where element addition takes place over $\mathbb{R}$.  For example, if we take the vectors that are the columns of the identity matrix as $S$ this will do.

What is the maximum number of vectors one can choose that has this
property?

I previously asked this question on MSE . An explicit construction of $17$ vectors was given by Oleg567 using computer search and an upper bound of $45$  was given by jpvee simply using the observation that $\sum_{k=1}^{17} {46 \choose k} > (17+1)^{10}$ implies that $46$ vectors is impossible.

Lower bound improved to $18$ by Oleg567. Upper bound still stuck at $45$ although it seems implausible the true value is far from the current lower bound.

Upper bound of $36$ given by Seva.

Conjecture Feb 24, 2014. I conjecture the optimal solution size  is  $\lfloor \frac{1}{2} (n+1) \log_2(n+1) \rfloor$. For $n=2\dots 15$ this is $2, 4, 5, 7, 9, 12, 14, 16, 19, 21, 24, 26, 29, 32$.

New lower bound of $19$ by Brendan McKay.

New upper bound of $30$ by Brendan McKay.
 A: Inspired by Seva's probabilistic method, I will show how to improve the upper bound to 30.  Imagine we have a 0-1 matrix $A$ of 31 rows and 10 columns. I will show that there are two different subsets of the rows that have the same sum.
Define the 10-dimensional random variable $X=(X_1,\ldots,X_{10})$ whose value is the sum of a random subset of the rows.  This is the same random variable defined by Seva.  Seva proceeded by showing that each  $X_j$ is concentrated on a few values. I will improve the bound by considering the components in pairs.
Write $X$ as $(Y_{12},Y_{34},Y_{56},Y_{78},Y_{9,10})$, where $Y_{12}=(X_1,X_2)$,  $Y_{34}=(X_3,X_4)$, and so forth.  The distribution of $Y_{12}$ depends only on the parameters $w_{01},w_{10},w_{11}$, which are respectively the number of times that 01, 10, 11 occur in the first two columns of $A$.  (And so 00 occurs $31-w_{01}-w_{10}-w_{11}$ times.)  The probability generating function for $Y_{12}$ is
$$ F_{12}(x_1,x_2) = 2^{-w_{01}-w_{10}-w_{11}}(1+x_1)^{w_{10}} (1+x_2)^{w_{01}} (1+x_1x_2)^{w_{11}}. $$
(The coefficient of $x_1^ax_2^b$ is the probability that $Y_{12}=(a,b)$.)
By trying all possible $w_{01},w_{10},w_{11}$, we find that in each case there is some set $K_{12}$ of 55 values such that 
$$\textrm{Prob}( Y_{12}\notin K_{12}) \le p = \frac{300387}{2097152} \approx 0.1432.$$
This calculation is easy for a computer: just expand $F_{12}$ and sum the largest 55 coefficients. One worst case is $w_{01}=w_{10}=10, w_{11}=11$.
By symmetry, there are also sets $K_{34},\ldots,K_{9,10}$ of size 55 containing at least the fraction $1-p$ of $Y_{34},\ldots,Y_{9,10}$, respectively. By the union bound, at least the fraction $1-5p$ of $Y$ lies in 
$$K = K_{12}\times \cdots\times K_{9,10}.$$
However, $(1-5p)2^{31} > |K| = 55^5$, Therefore, there are two values the same.
It should be possible to do better by grouping the columns even more. I think a non-trivial but plausible computation could handle the 10 columns in two groups of 5.
A: Following marshall's comment below, I (sadly) had to completely re-write my original answer.
A famous open conjecture of Paul Erdos, first stated about 80 years ago, is that if all subset sums of an integer set $S\subset[1,n]$ are pairwise distinct, then $|S|<\log_2n+O(1)$ as $n\to\infty$. (Here $\log_2$ denotes the base-$2$ logarithm.) In modern terms, a subset of an abelian group, all of whose subset sums are pairwise distinct, is called dissociated. Similarly to Erdos' original problem, one can ask how large can dissociated subsets of other "natural" sets in abelian groups be. Say, you are asking what is the largest possible size of a dissociated subset of the set $\{0,1\}^n\subset{\mathbb R}^n$, and this particular problem has been studied by a number of authors. It is known that the largest size of its dissociated subset is
  $$ \frac12(1+o(1))\,n\log_2 n; $$
see, for instance, this paper by Bshouty for details and a historical account.

Added 19.02.14 / Edited 24.02.14
A bug in my original post fixed, what I can show is that for $n=10$, at most
$36$ vectors can be found. Perhaps, with some effort this can be pushed a
little further to yield an even smaller bound. Here is the argument.
Assuming that $S=\{s_1,\ldots,s_m\}$ is a dissociated subset of $\{0,1\}^n$,
for each $i\in[m]$ and $k\in[n]$ write $s_i=(s_{i1},\ldots,s_{in})$ and
$w_k:=s_{1k}+\dotsb+s_{mk}$. Choose $\epsilon_1,\ldots,\epsilon_m\in\{0,1\}$
independently of each other and randomly with ${\mathsf
P}(\epsilon_i=0)={\mathsf P}(\epsilon_i=1)=1/2$, and let
$X_k:=\epsilon_1s_{1k}+\dotsb+\epsilon_ms_{mk}$ $(k\in[n])$; thus,
$X_1,\ldots,X_n$ are random variables with $X_k\sim B(w_k,1/2)$ and
$\epsilon_1s_1+\dotsb+\epsilon_ms_m=(X_1,\ldots,X_n)$.
Fix an integer $b\ge 0$ and for each $k\in[n]$, let $I_k$ denote the block of
$b$ consecutive integers, centered around $w_k/2$. (If $w_k$ and $b$ are of
the same parity, there is a unique such block, otherwise there are two
blocks.) Write $p_w(b)$ for the length $w+1-b$ tail of the binomial
distribution $B(w,1/2)$; that is, $p_w(b)$ is the probability that a random
variable with this distribution will not take one of its $b$ most probable
values. We then have ${\mathsf P}(X_k\notin I_k)=p_{w_k}(b)\le p_m(b)$ for
each $k\in[n]$; hence, by the union bound, $X_k\in I_k$ holds for all
$k\in[n]$ with probability at least $1-n\cdot p_m(b)$.
We now observe that $X_1\in I_1,\ldots,X_n\in I_n$ means that
$\epsilon_1s_1+\dotsb+\epsilon_ms_m\in I_1\times\dotsb\times I_n$, the
probability of which is $2^{-m}T$, where $T$ is the number of subsets sums of
$S$ (that is, the number of choices of $\epsilon_1,\ldots,\epsilon_m$) that
fall into the box $I_1\times\dotsb\times I_n$. However, since $S$ is
dissociated, the number of such subset sums does not exceed the total number
of integer points inside $I_1\times\dotsb\times I_n$, which is $b^n$. As a
result, we get
  $$ 2^{-m}b^n \ge 1-n\cdot p_m(b), \tag{$\ast$} $$
and to show that $m\le 36$ it remains to notice that ($\ast$) is invalid for
$n=10,\ m=37$, and $b=11$.
