You Only Have Their Sums

Let the numbers that the people are thinking of be denoted as $a_i$ . Each consecutive pair announces the value of $a_i + a_{i+1}$ .

Hence, we are given $n$ linear equations in $n$ variables, and are asked "When can we solve this system completely?". We know that $n$ equations is just enough to solve it in $n$ variables. Moreover, if equations are essentially duplicates, then the information that we obtained is not sufficient to solve uniquely.

When $n$ is even, we see that by taking the sum of $\frac{n}{2}$ pairs consisting of every other pair, we can find the sum of all of the people. This information is duplicated when we take the sum of the other $\frac{n}{2}$ pairs, so we have at least one wasted equation. This suggests that there are multiple solutions: If $( a_1, a_2, \ldots, a_n )$ is a solution to the linear system, then

$( a_1 + k, a_2 -k, a_3 + k, a_4 - k, \ldots, a_{n-1} +k , a_n -k )$

is also another solution (since each consecutive pair perfectly cancels). Hence, the solution isn't unique, so the outsider is not able to deduce all of the individual numbers.

When $n$ is odd, as it turns out we have exactly enough information to determine each individual number.
First, if we add up all the pairs and divide by 2, we get the sum of each individual value. Let's call this $A = \sum a_i$ .
Second, to find the value of $a_j$ , we just go round and subtract each other pair from him, namely $a_{j+1} + a_{j+2} , a_{j+3} + a_{j+4} , \ldots , a_{j-1 } + a_{j-2}$ . We can pair up these $n-1$ people perfectly.
Hence, the outside can use this approach to deduce all of the individual numbers.

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In linear algebra terms, we are saying that the system is not linearly independent .

We could arrive at the conclusion by studying the determinant of the matrix:

$A_n = \begin{pmatrix} 1 & 1 & 0 & \ldots & 0 & 0 \\ 0 & 1 & 1 & \ldots & 0 & 0 \\ & & & \ddots & & \\ 0 & 0 & 0 & \ldots & 1 & 1 \\ 1 & 0 & 0 & \ldots & 0 & 1 \\ \end{pmatrix}$

In particular, the above shows that $\det ( A_n) = 0$ if and only if $n$ is even.