[ Inequalities problems. Case analysis ]

A circular road is divided into kilometer segments by poles, and it is known that the number of poles is even. One of the poles is painted yellow, another is painted blue, and the rest are painted white. We will call the distance between poles the length of the shortest arc connecting them. Find the distance from the blue pole to the yellow one if the sum of the distances from the blue pole to the white ones is 2008 km.

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Let there be $-2n$ poles on a circular road. We calculate the sum of distances from the blue pole to all others: $2(1 + 2 + \ldots + (n-1)) + n = n(n-1) + n = n^2$. Therefore, $n^2 > 2008$. Since the distance from the blue pole to the yellow one does not exceed $n$, then $n^2 - n \leq 2008$, which means $n(n-1) \leq 2008$. Notice that $44^2 < 44 \cdot 45 < 2008 < 45^2 < 45 \cdot 46$. Therefore, the only natural number satisfying both inequalities is $n = 45$. Then $n^2 = 2025$, and the distance from the blue pole to the yellow one is $2025 - 2008 = 17$ km.

## Answer

17 km.