MOPSS, 14th March, 2026

We claim that Igor, the first player, has a winning strategy. Note that the numbers \(1, 2, 3, \dots, 30\) are congruent to \(31, 32, 33, \dots, 60\) modulo \(3\), respectively. Igor in his first turn places a \(-\) sign between \(30\) and \(31\). Next, after each turn of Ruslan, which is to place a sign in one of the two halves, that is, to place a sign between two consecutive numbers between \(1\) and \(30\) or to place a sign between two consecutive numbers between \(31\) and \(60\), Igor can respond by placing the corresponding sign in the other half. Here the corresponding sign means that if Ruslan places a \(\times\) sign, then Igor also places a \(\times\) sign, and if Ruslan places a \(+\) sign, then Igor places a \(-\) sign, and if Ruslan places a \(-\) sign, then Igor places a \(+\) sign. After all signs are placed, the resulting expression can be written as the difference of two expressions, one of which consists of the numbers \(1, 2, 3, \dots, 30\) and the other of which consists of the numbers \(31, 32, 33, \dots, 60\). The strategy of Igor ensures that these two expressions are congruent modulo \(3\). Therefore, Igor wins.

The first player can use the following strategy to win the game. Before the game starts, the first player calculates the sum of the numbers on the cards in the odd positions, and the sum of the numbers on the cards in the even positions. Since there are \(2006\) cards, one of these sums is greater than or equal to the other. If the sum of the numbers on the cards in the odd positions is greater than or equal to the sum of the numbers on the cards in the even positions, then the first player always takes the cards in the odd positions. Otherwise, the first player always takes the cards in the even positions. In either case, the first player wins. Note that the first player can always take the cards in the odd positions or the cards in the even positions, because the first player starts the game and there are an even number of cards. Indeed, if the first player takes the card in the odd (resp. even) position in the first turn from the left (resp. right) end of the row, then after each turn of the second player, the first player can respond by taking the card adjacent to the card taken by the second player, which is in the odd (resp. even) position. Thus, the first player can always take the cards in the odd positions or the cards in the even positions, and hence ensures that the sum of the numbers on the cards taken by the first player is greater than or equal to the sum of the numbers on the cards taken by the second player. Consequently, the first player wins.

We claim that Becky has a winning strategy if and only if \(n \equiv 2 \pmod{4}\). If \(n \equiv 2 \pmod{4}\), then compares the sum of the values of the stars in the odd positions, and the sum of the values of the stars in the even positions. Since \(n \equiv 2 \pmod{4}\), one of these sums is strictly greater than the other. If the sum of the values of the stars in the odd positions is greater than the sum of the values of the stars in the even positions, then Becky always takes the stars in the odd positions. Otherwise, Becky always takes the stars in the even positions. In either case, Becky wins.

If \(n = 2m + 1\) with \(m \geq 2\), then Anya places the stars with values \(1, 2, \dots, m + 1\) in the odd positions from left to right, and places the remaining stars in the even positions from left to right. Then, no matter how Becky plays, Anya takes the adjacent star, which is in an even position. Note that the sum of the values of the stars in the odd positions is \[ 1 + 2 + \dots + (m + 1) = \frac{(m + 1)(m + 2)}{2}, \] and the sum of the values of the stars in the even positions is \[ (m + 2) + (m + 3) + \dots + (2m + 1) = \frac{3m}{2}(m + 1) . \] Since \(m \geq 2\), we have \[ \frac{3m}{2}(m + 1) > \frac{(m + 1)(m + 2)}{2}, \] so Anya wins.

Assume that \(n = 4m + 4\) with \(m \geq 1\). Then Anya uses a similar strategy to force a draw or a win for Anya. First, Anya places the stars with values \(m + 1, m + 2, \dots, 2m\) in the first \(m\) odd positions from left to right, the stars with values \(2m + 1, 2m + 2, \dots, 3m\) in the next \(m\) odd positions from left to right, the stars with values \(3m + 1, 3m + 2, \dots, 4m\) in the first \(m\) even positions from left to right, and the stars with values \(1, 2, \dots, m\) in the next \(m\) even positions from left to right. In other words, Anya places the stars in the order

\(\begin{align*} & m + 1, 3m + 1, m + 2, 3m + 2, m + 3, 3m + 3,\dots, 2m, 4m,\\ & 2m + 1, 1, 2m + 2, 2, \dots, 3m, m \end{align*} \\)

from left to right. Anya play by taking the star adjacent to the star taken by Becky in each turn. We claim that, as a consequence of the above strategy, the game ends in a draw or a win for Anya.

To prove this claim, let us first consider the case when Becky takes more or an equal number of stars from the left end of the row than from the right end of the row. Note that Anya takes the stars with values \(3m + 1, 3m + 2, \dots, 4m\). As per Anya’s strategy,

\(\begin{align*} & \text{the sum of the values of stars taken by Anya} \\ & \quad - \text{the sum of the values of stars taken by Becky} \\ & \geq (3m + 1) + (3m + 2) + \dots + (4m) \\ & \quad - \left( (m + 1) + (m + 2) + \dots + (2m) \right) \\ & \quad - \vert (2m + 1) - 1 \vert - \vert (2m + 2) - 2 \vert - \dots - \vert 3m - m \vert \\ & = 2m^2 - 2m^2 \\ & = 0. \end{align*} \\)

Next, we consider the case when Becky takes more stars from the right end of the row than from the left end of the row. Note that Anya takes the stars with values \(2m + 1, 2m + 2, \dots, 3m\). As per Anya’s strategy,

\(\begin{align*} & \text{the sum of the values of stars taken by Anya} \\ & \quad - \text{the sum of the values of stars taken by Becky} \\ & \geq (2m + 1) + (2m + 2) + \dots + (3m) \\ & \quad - \left( 1 + 2 + \dots + m \right) \\ & \quad - \vert (3m + 1) - (m + 1) \vert - \vert (3m + 2) - (m + 2) \vert - \dots - \vert 4m - 2m \vert \\ & = 2m^2 - 2m^2 \\ & = 0. \end{align*} \\)

This proves the previous claim.

This establishes that Becky has a winning strategy if and only if \(n \equiv 2 \pmod{4}\).

In the following, we assume that \(n \neq 1\). Using division algorithm, for any two distinct positive integers \(a\) and \(b\), it follows that \[ \mathrm{gcd} (a, b) \leq a \leq \frac{b}{2} \] if \(a\) divides \(b\), and \[ \mathrm{gcd} (a, b) \leq \frac{a}{2} \] if \(a\) does not divide \(b\), and hence \(\mathrm{gcd} (a, b) \leq \frac{1}{2} \max(a, b)\). This implies that the GCD of any two adjacent numbers in the row is at most \(n/2\). Therefore, the maximum possible number of distinct GCDs obtained, is at most \(\lfloor n/2 \rfloor\). We will show that this upper bound can be achieved by the following arrangement of the numbers \(1, 2, \ldots, n\). For each odd positive integer \(k \leq n\), write the integers between \(1\) and \(n\) in a block \(\mathcal{B}_k\), having \(k\) as its largest odd divisor, in an increasing order, that is, the elements of \(\mathcal{B}_k\) are put in the order \[ k, 2k, 4k, \ldots, 2^{e_k} k , \] where \(e_k\) is the largest non-negative integer such that \(2^{e_k} k \leq n\). Next, we put the blocks \(\mathcal{B}_k\) in a row as \(k\) increases across the odd positive integers less than or equal to \(n\).

Claim. Let \(1 \leq m \leq n/2\) be a positive integer. Then \(m\) is equal to the GCD of some pair of adjacent numbers in the above arrangement.

Proof of the claim. Since \(m\) is at most \(n/2\), it follows that \(2m\) is at most \(n\). If \(k\) denotes the largest odd divisor of \(m\), then \(k\) is less than or equal to \(m\) and hence \(k \leq n\). Hence, in the above arrangement, the block \(\mathcal{B}_k\) has appeared, and the numbers \(m\) and \(2m\) are adjacent in this block. Note that the GCD of \(m\) and \(2m\) is \(m\). This proves the claim.

Therefore, the GCDs of the adjacent numbers in the above arrangement include all the integers \(1, 2, \ldots, \lfloor n/2 \rfloor\). Since the number of distinct GCDs obtained is at most \(\lfloor n/2 \rfloor\), it follows that the maximum possible number of distinct GCDs obtained is \(\lfloor n/2 \rfloor\).