Croatia_2018

Let $a$, $p$, $q$ and $b$ be four consecutive numbers in an alternating arrangement. Assume $p>q$. Then $a<p$ and $q<b$.

The pair $(p,q)$ is not good if and only if $a>b$. Therefore, we conclude that $(p,q)$ is not good if and only if $p$ is the largest, and $q$ the smallest number of the quadruple $(a,p,q,b)$.

If $(p,q)$ is not good, then $(a,p)$ is, because $a$ is greater than or equal to $q$. Furthermore, $(q,b)$ is good because $b$ is not greater than $p$.

This shows that, of any two pairs sharing an element, at least one is a good pair. We conclude that there are at least $150$ good pairs.

Assume that there is an alternating arrangement $a_1,\dots ,a_{300}$ in which there are exactly $150$ good pairs.

Without loss of generality, we can assume that $a_1>a_2$, and that $(a_1,a_2)$ is not a good pair. Then $(a_{2k},a_{2k+1})$ are both good, while the pair $(a_{2k-1},a_{2k})$ is not, for any $k=1,2,\dots ,149$. We also notice that $(a_{299},a_{300})$ cannot be a good pair.

Since $(a_1,a_2)$ is not good, we conclude that $a_1>a_3$. Similarly, since $(a_3,a_4)$ is not good, we get $a_3>a_5$. Continuing this way, we get $$a_1>a_3>a_5>\cdots >a_{299}>a_1,$$ thus arriving at a contradiction. This shows that it is impossible to have exactly $150$ good pairs. Therefore, there must be at least $151$ of them.

An example of an alternating arrangement containing exactly $151$ good pairs is $3,2,5,4,7,6,\dots ,299,298,300,1$.