Cesko-Slovacko-Poljsko

First we prove the numbers $a_1, \dots, a_n$ are mutually relatively prime. If this is not true, we have $(a_i, a_j) = d > 1$ for some $i \ne j$. Put $a_i = u \cdot d$, $a_j = v \cdot d$. Set $b_i = 1$, $b_j = 2$. By (ii), there exist $m$, $c_i$, and $c_j$ such that
$$m \cdot b_i = c_i^{a_i} \quad \text{and} \quad m \cdot b_j = c_j^{a_j}, \quad \text{therefore} \quad m = (c_i^{a_i})^d \quad \text{and} \quad 2m = (c_j^{v_j})^d.$$
So $2(c_i^{a_i})^d = (c_j^{v_j})^d$, which is not possible, since the exponent of $2$ in the prime factorization of the right hand side is and the one of the left hand side is not a multiple of $d$.

Assume that $a_1, \dots, a_n$ are mutually relatively prime. We shall prove that the condition (ii) is fulfilled. Let $b_1, \dots, b_n$ be any $n$-tuple of positive integers and $p_1, \dots, p_k$ be all the prime divisors of $b_1, \dots, b_n$. We look for $a_i$ in the form
$$m = p_1^{a_1} \cdots p_k^{a_k}.$$
For $i = 1, \dots, n$, denote by $\beta_{i,j}$ the exponent of $p_j$ in the prime factorization of $b_i$. In order for $m \cdot b_i$ to be an $a_i$-th power, it suffices $\alpha_j + \beta_{i,j}$ to be a multiple of $a_i$ for $j = 1, \dots, k$. So we need $\alpha_j$ fulfilling the congruences
$$\alpha_j \equiv -\beta_{1,j} \pmod{a_1}, \quad \alpha_j \equiv -\beta_{2,j} \pmod{a_2}, \quad \dots \quad \alpha_j \equiv -\beta_{n,j} \pmod{a_n}.$$
The existence of such $\alpha_j$ is guaranteed by Chinese Remainder Theorem (as $a_1, \dots, a_n$ being mutually relatively prime).

We have proved the condition (ii) is satisfied if and only if the numbers $a_1, \dots, a_n$ are mutually relatively prime. Among $1, 2, \dots, 50$, there are exactly $15$ primes. If $n \ge 17$ then among $2 \le a_2 < a_3 < \dots < a_n \le 50$, there must be at least two numbers sharing the same prime number in their prime factorizations, hence not being relatively prime. Therefore we have $n \le 16$.

If $n = 16$, then $a_1 = 1$ and $a_2, a_3, \dots, a_{16}$ must be the powers of different primes. We list the powers of primes which can be used:
$$\begin{array}{lclcl} p & = & 2: & 2, 4, 8, 16, 32, \\ p & = & 3: & 3, 9, 27, \\ p & = & 5: & 5, 25, \\ p & = & 7: & 7, 49, \\ p & \ge & 11: & \text{only } p. \end{array}$$
Then the total number of the corresponding $16$-tuples is $5 \cdot 3 \cdot 2 \cdot 2 = 60$.