NMO Selection Tests

a) Two real numbers (not necessarily distinct) always form a suitable collection. Let $n \ge 3$ and consider the initial collection. Note that any $a$ can be paired off with some $b \neq a$ to form an eligible pair: Otherwise, $a+b=s+1=a+c$ for distinct $b, c \neq a$, so $b=c$, contradicting the fact that the initial numbers are pairwise distinct. Hence the initial collection is suitable.

Choose an eligible pair $a_1, a_2$ and use the formula to replace them by $b_1$. As the remaining numbers are pairwise distinct, there is at most one that cannot be paired with $b_1$ to form an eligible pair, so there are at least $n-2$ candidates to form an eligible pair with $b_1$. Let $a_3$ be one such and use the formula to replace $b_1$ and $a_3$ by $b_2$. Repeat the argument to replace $b_2$ and some $a_4$ by $b_3$ and so on and so forth all the way down to some $b_{n-2}$ and $a_n$ (possibly, $b_{n-2} = a_n$). These latter form an eligible pair, so they can be replaced by a single number.

b) Let $c_1, c_2, \dots, c_n$ be the initial numbers. The final number is $\sum_i c_i + \sum_{i<j} c_i c_j$. This is clearly the case if $n=2$, so let $n \ge 3$.

Consider a generic stage $x_1, x_2, \dots, x_m$ and let $s = x_1 + x_2 + \dots + x_m$. We will prove that $s + \sum_{1 \le i < j \le m} x_i x_j$ does not change upon passing to the next stage. Let $x$ be the number obtained by replacing an eligible pair $(x_k, x_\ell)$. Then $s$ changes by $x - x_k - x_\ell$ and the other sum changes by $-x_k x_\ell - (x_k + x_\ell)(s - x_k - x_\ell) + x(s - x_k - x_\ell)$, so the overall change is
$$x - x_k - x_\ell - x_k x_\ell - (x_k + x_\ell)(s - x_k - x_\ell) + x(s - x_k - x_\ell).$$
Finally, express $x$ in terms of $s$, $x_k$ and $x_\ell$ and carry out calculations to show that the overall change vanishes, whence the desired invariance.