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 $$m\cdot b_i=c_i^{a_i}\text{and}m\cdot b_j=c_j^{a_j},\text{therefore}m=(c_i^{a_i}{)}^d\text{and}2m=(c_j^{v_j}{)}^d.$$ 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 $${\alpha }_j\equiv -{\beta }_{1,j}(\mathrm{mod}{a}_1),{\alpha }_j\equiv -{\beta }_{2,j}(\mathrm{mod}{a}_2),\dots {\alpha }_j\equiv -{\beta }_{n,j}(\mathrm{mod}{a}_n).$$ 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 $2\le a_2<a_3<\cdots <a_n\le 50$, 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: $$\begin{array}{lclc}p& =& 2:& 2,4,8,16,32,\\ p& =& 3:& 3,9,27,\\ p& =& 5:& 5,25,\\ p& =& 7:& 7,49,\\ p& \ge & 11:& \text{only }p.\end{array}$$ Then the total number of the corresponding $16$-tuples is $5\cdot 3\cdot 2\cdot 2=60$.