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
$$a_1 > a_3 > a_5 > \dots > a_{299} > a_1,$$
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
$3, 2, 5, 4, 7, 6, \ldots, 299, 298, 300, 1$.