Belarusian Mathematical Olympiad

Answer: $n = 2$.

Note that if $n = 2$, the sequence has the form $a_k = k(k+1)$ since consecutive numbers are always coprime. It is clear that $k(k+1) < (k+1)(k+2)$, so the sequence is increasing.

First solution. Let us show that if $n \geq 3$ the sequence is not increasing from any number. Choose $k = np$ where $p$ is an arbitrary prime number greater than $n$. All numbers $np+1, np+2, \dots, np+n-1$ are not divisible by $p$ and there is at least one even among them. The number $n(p+1)$ is also not divisible by $p$ and $p+1$ is even. Hence
$$a_k = p \cdot \text{lcm}(n, np+1, np+2, \dots, np+n-1), \\ a_{k+1} \le \frac{p+1}{2} \cdot \text{lcm}(n, np+1, np+2, \dots, np+n-1).$$
Therefore $a_k > a_{k+1}$ for $k$ large enough and the sequence $a_k$ is not increasing from any number.

Second solution. We will show how else one can prove that for $n \geq 3$ the sequence is not increasing from any moment. Suppose that $a_k$ is increasing from number $k_0$ then for all $k \geq k_0$ holds $a_{k+1} \geq a_k$. Consider $k = m! - n$ where $m > \max(n! + n, k_0)$.

Denote $\text{lcm}(k+1, \dots, k+n-1)$ by $N$. Then the inequality $a_{k+1} \ge a_k$ is equivalent to $\text{lcm}(N, m!) > \text{lcm}(m! - n, N)$. Using the well known equality $\text{lcm}(a, b) \cdot \text{gcd}(a, b) = a \cdot b$ we obtain the equivalence:
$$\text{lcm}(N, m!) > \text{lcm}(m! - n, N) \iff \frac{N \cdot m!}{\text{gcd}(N, m!)} > \frac{(m! - n) \cdot N}{\text{gcd}(m! - n, N)},$$
or, after equivalent transformations,
$$n \cdot \text{gcd}(N, m!) > m! \cdot (\text{gcd}(N, m!) - \text{gcd}(m! - n, N)). \quad (1)$$
It is clear that $\text{gcd}(m!, m! - l) = \text{gcd}(m!, l) = l$ for all $l$ from $1$ to $n-1$, hence $(n-1)! \ge \text{gcd}(N, m!) \ge \text{lcm}(1, 2, \dots, n-1)$. Moreover, since $m!$ is divisible by $l$, the number $\text{gcd}(m! - n, m! - n + l) = \text{gcd}(n, n-l)$ is a divisor of $n$. So the number $\text{gcd}(m! - n, N)$ is a divisor of $n$ too. Indeed $\nu_p(N) = \max(\nu_p(m! - n + 1), \dots, \nu_p(m! - n + (n-1)))$ for any prime divisor $p$ of $N$.
From obtained inequalities and (1) it follows that
$$n \cdot (n-1)! > m! \cdot (\text{lcm}(1, 2, \dots, n-1) - n). \quad (2)$$
Since $\text{lcm}(1, 2, \dots, n-1) \ge (n-2)(n-1)$, for $n \ge 4$ the right hand side of (2) is not less than $m!$ whence $n! > m!$, a contradiction.
If $n=3$ then $\text{gcd}(N, m!) = \text{gcd}(m! - 2, m!) = 2$. But
$$\text{gcd}(m! - n, N) = \text{gcd}(m! - 3, (m! - 2)(m! - 1)) = 1,$$
so the inequality (1) will become $6 > m!$ which is wrong.