Lemma. Let $x, y, z$ be integers with $1 \le x, y, z \le N$. Then at least one of $(x, y, z)$, $(y, z, x)$, or $(z, x, y)$ belongs to $S$.

proof. Consider a sequence $a_1 = x, a_2 = y, a_3 = z, a_4 = x, a_5 = y, a_6 = z, \dots$. Then, for every positive integer $i$, $(a_i, a_{i+1}, a_{i+2})$ is one of $(x, y, z)$, $(y, z, x)$, or $(z, x, y)$. Therefore, by the assumption of the problem, at least one of these triples must belong to $S$. ■

Consider the set
$$X = \{(x_1, y_1, z_1), (x_2, y_2, z_2), \dots, (x_n, y_n, z_n), (y_1, z_1, x_1), (y_2, z_2, x_2), \dots, (y_n, z_n, x_n), \\ (z_1, x_1, y_1), (z_2, x_2, y_2), \dots, (z_n, x_n, y_n)\}.$$
Let $p, q, r$ be integers with $1 \le p, q, r \le N$. Applying Lemma to $(x, y, z) = (p, q, r)$, it follows that at least one of $(p, q, r)$, $(q, r, p)$, or $(r, p, q)$ belongs to $S$, say it is $(x_j, y_j, z_j)$ for some $1 \le j \le n$. Then, $(p, q, r)$ is equal to one of $(x_j, y_j, z_j)$, $(z_j, x_j, y_j)$, or $(y_j, z_j, x_j)$, and therefore, $(p, q, r)$ belongs to $X$. Thus, it has been shown that $X$ is the set of all $N^3$ triples $(p, q, r)$ with $1 \le p, q, r \le N$.
By applying Lemma to $(x, y, z) = (1, 1, 1), (2, 2, 2), \dots, (N, N, N)$, we find that these triples $(1, 1, 1), (2, 2, 2), \dots, (N, N, N)$ all belong to $S$. If $(x_j, y_j, z_j)$ is equal to one of them, we have $(x_j, y_j, z_j) = (y_j, z_j, x_j) = (z_j, x_j, y_j)$, hence the number of elements of $X$ is less than or equal to $3n - 2N$. Therefore we have $3n - 2N \ge N^3$, thus $n \ge \frac{N^3 + 2N}{3}$.

Next, we construct a set $S$ satisfying $n = \frac{N^3 + 2N}{3}$. Define $S$ to be the set of triples $(x, y, z)$ with $1 \le x, y, z \le N$ such that either
* $x > y$ and $x \ge z$, or

$$\bullet x = y = z.$$
If we fix an integer $x$ with $1 \le x \le N$, the number of pairs $(y, z)$ such that $(x, y, z)$ belongs to $S$ is equal to $x(x-1)+1$. Therefore, the number of elements in $S$, denoted by $n$, is
$$n = \sum_{x=1}^{N} (x(x-1)+1) = \frac{N^3 + 2N}{3}.$$
Assume, for the sake of contradiction, that there exists a sequence $a_1, a_2, \dots$ of integers $1, 2, \dots, N$ such that none of the triples $(a_1, a_2, a_3), (a_2, a_3, a_4), \dots$ belongs to $S$. Let $a_k$ be one of the largest terms among $a_1, a_2, \dots$. Then we have $a_k \ge a_{k+1}$ and $a_k \ge a_{k+2}$. Since $(a_k, a_{k+1}, a_{k+2})$ does not belong to $S$, we have $a_{k+1} = a_k$. Therefore, $a_{k+1}$ is also one of the largest terms in the sequence. Applying the same argument to $(a_{k+1}, a_{k+2}, a_{k+3})$, we obtain $a_{k+2} = a_{k+1}$. Hence, $a_k = a_{k+1} = a_{k+2}$, so $(a_k, a_{k+1}, a_{k+2})$ belongs to $S$, a contradiction. Therefore, this set $S$ satisfies the required condition.

This shows that the minimum value of $n$ is $\frac{N^3 + 2N}{3} = 333400$.