$\underline{\text { Senderov V.A. }}$

A thousand non-overlapping arcs are placed on a circle, and two natural numbers are written on each of them. The sum of the numbers on each arc is divisible by the product of the numbers on the next arc in the clockwise direction. What is the greatest possible value of the largest of the written numbers?
Let's prove that the numbers on the circle do not exceed 2001.

Lemma 1. Let \( x \) and \( y \) be natural numbers. If \( xy = x + y \), then \( x = y = 2 \), and if \( x > 0, y > 0, x \leq y \), then the sum \( x + y \) decreases as \( x \) increases.

Proof. This follows from the decrease of the function \( f(x) = x + \frac{c}{x} = \left(\sqrt{\frac{c}{x}} - x\right)^2 + 2\sqrt{c} \), where \( c > 0 \), on the interval \( (0, \sqrt{c}) \).

Divide the sum of the numbers in each pair by the product of the numbers in the next (clockwise) pair and multiply the resulting quotients. By the condition, we get an integer. On the other hand, this product is the product of numbers of the form \( \frac{a+b}{ab} \). From this and Lemma 1, it follows that if at least one pair is different from (2, 2), then there will be a pair of the form (1, k). Start with this pair and move along the circle clockwise. First case. Subsequent pairs have the form (1, k+1), (1, k+2), ..., (1, k+999). Hence, \( k + 1000 \) is divisible by \( k \), from which 1000 is divisible by \( k \). But then \( k \leq 1000 \) and \( k + 999 \leq 1999 \).

Second case. There will be a pair of the form (1, l) such that the next pair \((a, b)\) is different from (1, l+1). By the condition, \( ab \) is a divisor of the number \( s_1 = 1 + l \). If

\( ab = l + 1 \), then by Lemma 2, \( s_2 = a + b \leq 2 + \frac{1}{2}(l + 1) \).

If \( ab \leq \frac{1}{2}(l + 1) \), then by Lemma 2, \( a + b \leq 1 + \frac{1}{2}(l + 1) \). Therefore, in this case, inequality (*) also holds. For the next pair \( (c, d) \), Lemma 2 gives \( s_3 = c + d \leq cd + 1 \leq 3 + \frac{1}{2}(l + 1) \), and so on. For the sum \( s_{1000} \) of the numbers in the pair preceding \((1, l)\), we get

\( s_{1000} \leq 1000 + \frac{1}{2}(l + 1) \).

On the other hand, \( s_{1000} \) is divisible by \( l \), so \( s_{1000} \geq l \). The last two inequalities give \( l \leq 2001 \). Then from the above estimates, it follows that for any pair different from (1, l), the inequality \( s \leq 2001 \) holds. Therefore, each number in these pairs does not exceed 2000.

Thus, the estimate of 2001 is proven. From the reasoning, it is easy to obtain an example: \((1, 2001), (2, 1001), (1, 1003), (1, 1004), (1, 1005), \ldots, (1, 2000)\).

## Answer

2001.