Iranian Mathematical Olympiad

a) Let $a_1 \le a_2 \le \dots \le a_n$ be the stolen numbers. Since all the partial sums are positive, all the stolen numbers must be positive. Obviously $a_1$ is the smallest number among the partial sums. Suppose that numbers $a_1, a_2, \dots, a_i$ have been determined. Omit all the partial sums of $a_1, a_2, \dots, a_i$ from the partial sums of $a_1, a_2, \dots, a_n$. The smallest number among the remaining numbers, must be $a_{i+1}$, so all the numbers will be determined uniquely.

b) Suppose that $a_1 \le \dots \le a_k < 0 \le a_{k+1} \le \dots \le a_n$ are the stolen numbers and $s_1 \le s_2 \le \dots \le s_{2^n-1}$ are the partial sums. Note that if $s_1 > 0$, the problem is already solved in part (a). Therefore, we can assume that $s_1 < 0$. We have
$$(1 + x^{a_1})(1 + x^{a_2})\cdots(1 + x^{a_n}) = 1 + x^{s_1} + x^{s_2} + \cdots + x^{s_{2^n-1}}.$$
Obviously, $s_1$ is the sum of negative numbers $a_1, a_2, \dots, a_k$. Multiplying the above equation by $x^{-s_1}$ implies
$$(1+x^{-a_1})\cdots(1+x^{-a_k})(1+x^{a_{k+1}})\cdots(1+x^{a_n}) = x^{-s_1}+x^{s_1-s_1}+x^{s_2-s_1}+\cdots+x^{s_{2^n-1}-s_1}.$$
Hence we can say that we have the partial sums of stolen numbers $|a_1|, |a_2|, \dots, |a_n|$, and by part a, it is possible to find them uniquely.
Finally, we must show that $x^{s_1}$ can be uniquely written as a product of $x^{a_i}$'s. If there were two ways to do this, we would obtain a partial sum of $a_i$'s equal to 0, which leads to a contradiction.

c) The partial sums of two sets $\{1, 2, -3\}$ and $\{-1, -2, 3\}$ are the same, so adding 1389 0's to these sets will not change the partial sums.