Anybody solve it?

For each prime $p$, let $N_{p}$ be the number of elements of $M$ with $p$ dividing the denominator, and let $Z_{p}$ be the number of elements of $M$ for which $p$ dvides neither the numerator nor the denominator. If $|N_{p}| \geq 1$, then we must have $|N_{p}|+|Z_{p}| < 7$, and $-|N_{p}|+(7-|N_{p}|-|Z_{p}| \geq 0$ $2 |N_{p}|+|Z_{p}| \leq 7$ $|N_{p}|+|Z_{p}| \leq 6$. So if we let $k$ be the number of primes that appear in the denominator of any of the elements of $M$, for each of the $k$ primes there are at most 6 elements for which the prime does not divide the numerator; hence, letting $f(n)$ be the maximum cardinality of $M$ for a specific value of $n$, $f(n) \leq 6k+2^{n-k}, 0 \leq k \leq n$. Note that the above is a convex function of $k$, so the function is maximized when $k = 0$ or $k = n$. This gives $f(n) \leq 2^{n}$ and $f(n) \leq 6n$ respectively; for $n \geq 5$, $6n \leq 2^{n}$, so $f(n) \leq 2^{n}$. But we can achieve $2^{n}$ elements by simply letting $M$ be all the integer products of the $n$ primes. Ao $f(n) = 2^{n}$ for $n \geq 5$. The cases $n = 3$ and $n = 4$ require a more detailed bound. Summing the inequalities $2 |N_{p}|+|Z_{p}| \leq 7$ over all primes $p$ with $|N_{p}| \geq 1$, we get $2 \sum_{p \in S}|N_{p}|+\sum_{p \in S}|Z_{p}| \leq 7k$ (1) where $S$ is the set of primes that appear in the denominators of the elements of $M$. Now, for any element $m$ of $M$, $m$ is of the form $m = \prod_{i=1}^{a}p_{i}^{1}\prod_{j=1}^{b}q_{j}^{0}\prod_{k=1}^{c}r_{k}^{-1}$, and for each prime $q_{j}$, $m$ lies in $Z_{q_{j}}$, and for each prime $r_{k}$, $m$ lies in $Z_{r_{k}}$. So $m$ contributes $b+2c$ to the left and side of equation (1). Set $g(m) = 2b+c$. There are $2^{n-k}$ elements with $g(m) = 0$, $2^{n-k}k$ elements with $g(m) = 1$, and $2^{n-k}(k+\binom{k}{2})$ elements with $g(m) = 2$. For $n = 4$, we have that $6k+2{n-k}\leq 16$ for $0 \leq k \leq 2$. Let $h(i)$ be the sum of the smallest $i$ values of $g(m)$. For $k = 3$, we have $2, 6, 12$ elements with $g(m) = 0, 1, 2$, so $h(16) = 2(0)+6(1)+8(2) = 22 > 21$. So $M$ can have most 15 elements in this case. For $k = 4$, we have $1, 4, 10$ elements with $g(m) = 0, 1, 2$, so $h(17) = 1(0)+4(1)+10(2)+2(3) = 30 > 28$. So we can have at most 16 elements, and $f(4) = 16$. Now for $n = 3$. For $k = 1$, we have $h(8) = 4(0)+4(1) = 8 > 7$. For $k = 2$, we have $h(12) = 2(0)+4(1)+6(2) = 16 > 14$. For $k = 3$, we have $h(13) = 1(0)+3(1)+6(2)+3(3) = 24 > 21$. So the number of elements cannot be more that 12. We can achieve this with ${M = \{2, 3, 5, 6, 10, 15, 30, 15/2, 10/3, 6/5, 3/2, 2/3}$. Just to finish things off, obviously $f(1) = 3$; for $n = 2$, we cannot take all 9 possible elements, but removing $1/6$ is sufficient, so $f(2) = 8$.

Hint. Every prime $p_{k}$ is maximum in 3 denominators. Thus, $|N|\leq 3n$. Hint. On the competition many good participants got result $\left\lfloor \frac{8n}3 \right\rfloor$ which is NOT the answer.

Okay, it looks like you are requiring that the numbers are non-integral. The formula $2 |N_{p}|+|Z_{p}| \leq 7$ that I derived above shows that $|N_{p}| \leq 3$. Suppose first that not all primes appear in the denominator. Then $|N| \leq 3n-3$. I claim this is in fact achievable. Let $P = \prod_{i = 1}^{k}p_{i}$. For each $p_{k}$, $k = 1,2, \ldots, n-2$, let $S_{k}= \{\frac{P}{p_{k}^{2}}, \frac{P}{p_{k}^{2}p_{k+1}}, \frac{P}{p_{k}^{2}p_{k+1}p_{n}}\}$ and $S_{n-1}= \{\frac{P}{p_{n-1}^{2}}, \frac{P}{p_{n-1}^{2}p_{1}}, \frac{P}{p_{n-1}^{2}p_{1}p_{n}}\}$. Then $M = \bigcup_{i=1}^{n-1}S_{i}$ satisfies the requirement. If all primes appear in the denominator, then let $S_{1}, S_{2},$ and $S_{3}$ be the number of primes that appear in $1,2,$ and $3$ denominators respectively, and let $n_{i}= |S_{i}|, 1 \leq i \leq 3$. If $p$ is in $S_{3}$, then for at least two of the elements with $p$ in the denominator, there is another prime $q$ such that $q$ does not divide the numerator. On the other hand, there is at most one other element for which $p$ does not divide the numerator. (from $2 |N_{p}|+|Z_{p}| \leq 7$) If $p$ is in $S_{2}$, then there is at least one element for which a prime $q$ does not divide the numerator, and at most 3 elements for which $p$ does not divide the numerator. If $p$ is in $S_{1}$, there is at most 5 elements for which $p$ does not divide the numerator. Thus we have $n_{3}+3 n_{2}+5 n_{1}\geq 2n_{3}+n_{2}$ $2 n_{2}+5 n_{1}\geq n_{3}$ $3 n_{2}+6 n_{1}\geq n$. Thus $|N| = 3n_{3}+2n_{2}+n_{1}= 3n-n_{2}-2n_{1}\leq 3n-\frac{n}{3}= \frac{8n}{3}$. Again, this bound is achievable for $n \geq 4$. Let $m = \lfloor \frac{2n}{3}\rfloor$. For $k = 1,2, \ldots, m-1$, let $S_{k}= \{\frac{P}{p_{k}^{2}}, \frac{P}{p_{k}^{2}p_{k+1}}, \frac{P}{p_{k}^{2}p_{m+\rceil \frac{k}{2}\lceil}}\}$ and $S_{m}= \{\frac{P}{p_{m}^{2}}, \frac{P}{p_{m}^{2}p_{1}}, \frac{P}{p_{m}^{2}p_{m+\rceil \frac{m}{2}\lceil}}\}$. For $k = m+1, m+2, \ldots, n-1$, $S_{k}= \{\frac{P}{p_{k}^{2}}, \frac{P}{p_{k}^{2}p_{k+1}}\}$ and $S_{n}= \{ \frac{P}{p_{n}^{2}}, \frac{P}{p_{n}^{2}p_{m+1}}\}$. This satisfies the required condition. For $n \geq 7$, we have $3n-3 \geq \lfloor \frac{8n}{3}\rfloor$. So $f(n) = 3n-3$ for $n \geq 7$, and $f(n) =\lfloor \frac{8n}{3}\rfloor$ for $4 \leq n \leq 9$. For $n=3$, we can let $M = \{3/2, 5/2, 15/2, 2/3, 5/3, 10/3, 6/5 \}$. This is optimal; to do better, we must have $n_{3}= 2$ and $n_{2}= 1$. But for $p, q$ in $S_{3}$, we have $p$ appearing in three denominators, but at most twice in numerators of elements with $q$ in the denominator. Similarly for $q$. So the product of the six elements has $p$ and $q$ in the denominator, so any remaining elements must have both $p$ and $q$ in the numerator. That leaves room for only one other element, $\frac{pq}{r}$. So seven elements is maximal.

absolutely correct