More unfinished lines.

Amazing problem! After months of inactivity solving this was a great way to start again.

For notation's sake, I'll redefine some variables. $\triangle_i = a_i, \bigcirc = N, \square_i = b_i = -1,0,1$ . If $\displaystyle S = \sum_{1 \le i \le 16} b_i a_i$ is divisible by $N$ , $S \equiv 0 \text{ mod } N$ . Hence, I'll be working in $\text{mod }N$ . It should be noted that when working with $\text{mod }N$ , it is susficient to consider $a_i$ ranging from $0$ to $N-1$ .

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Ok here's my solution:

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Claim: For $N > 2^{16} - 1$ , there exists a set of integers $a_i$ such that no sequence of $b_i$ exists such that $S \equiv 0 \text{ mod } N$

Proof: Consider $a_i = 2^{i-1}, 1 \le i \le 16$ . Then every possible sequence $c_i = 0,1$ would lead to $\displaystyle S' = \sum_{1 \le i \le 16} c_i a_i$ being a unique integer from $1 \le S' \le 2^{16} - 1$ . This can be seen from the binary representation of $S'$ . Note that our choice of $a_i$ forces $\displaystyle -2^{16} + 1 \le S = \sum_{1 \le i \le 16} b_i a_i \le 2^{16} - 1$ . In order for $S \equiv 0 \text{ mod } N$ , $S = 0$ .

Now, a sequence $b_i$ can be computed by an element wise subtraction of 2 sequences of $c_i: c_i', c_i''$ : $b_i = c_i' - c_i''$ . Suppose $\displaystyle S_0' = \sum_{1 \le i \le 16} c_i' a_i, S_1' = \sum_{1 \le i \le 16} c_i'' a_i$ . Then $\displaystyle S = \sum_{1 \le i \le 16} b_i a_i = S_0' - S_1'$ . In order for $S = 0$ , $S_0' = S_1'$ . Since if $c_i' \neq c_i''$ , $S_0' \neq S_1'$ , the only possible way $S_0' = S_1'$ is if $c_i' = c_i''$ . This would mean $b_i = 0$ for all $1 \le i \le 16$ , which isn't allowed.

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This would mean that $N \le 2^{16} - 1$ . Here I make a bold claim:

Claim: $N = 2^{16} - 1$ .

Proof: If $N = 2^{16} - 1$ , then $S \text{ mod } N$ has $2^{16}$ possible values, $0 \le S \text{ mod } N \le 2^{16} - 1$ . Now we have 2 cases:

Case 1: $S = 0 \text{ mod } N$ . If this is so, then we are done! (Claim proven)

Case 2: $S \neq 0 \text{ mod } N$ . If this is so, then $S \text{ mod } N$ can have $2^{16} - 1$ possible values, $1 \le S \text{ mod} N \le 2^{16}-1$ . Now, each sequence $c_i$ would map to a (not necessarily unique) $\displaystyle S' = \sum_{1 \le i \le 16} c_i a_i$ . Since there are a total of $2^{16}$ possible sequences $c_i$ , by pigeonhole principle, there will be at least $\text{ceil}\left(\frac{2^{16}}{2^{16} - 1}\right) = 2$ , $c_i: c_i', c_i''$ such that $S_0' \equiv S_1' \text{ mod } N$ . As such, $b_i$ can simply be constructed by taking the element-wise subtraction of the $c_i', c_i''$ : $b_i = c_i' - c_i''$ . As such, there would exist a sequence $b_i$ such that $S = 0 \text{ mod } N$ -- a contradiction. A such, case 2 is impossible.

Hence we have proven that $N = 2^{16} - 1 = \boxed{65535}$ .