These numbers are exactly the primes $m$.

If $m=2n$ or $m=3n$ we get $n|1$ which is impossible. ( $m=2,3$ as obvious cases ) So in other cases assume $m$ isn't a multiple of $2$ or $3$ but isn't prime. $m=(2a+1)*q$ and take $n=aq$ to get $q\not | \binom{aq}{q}= a \cdot \frac{(aq-1)(aq-2)\cdots(aq-q+1)}{(q-1)!}$ If $m$ is prime, we get $n| \binom{n}{m-2n}= \frac{n}{3n-m} \cdot \binom{n-1}{m-2n}$ and $gcd(n,3n-m)=1$ as $gcd(n,m)=1$.

If $m$ is prime, it suffices to prove that if $(a,b)=1$ and $1 \le b \le a-1$ then $a | \binom{a}{b}$. To prove this, we use the Lagrange formula, $v_p(n!)=\frac{n-s_p(n)}{p-1}$ where $v_p(x)$ and $s_p(x)$ are the exponents of the maximum power of $p$ that divides $x$ and the sum of digits of $x$ in base $p$ respectively. It reduces to proving that $s_p(a-1)+1 \le s_p(b)+s_p(a-b)$. If $p \not | a$ then this is clear since $s_p(b)+s_p(a-b) \ge s_p(a)=s_p(a-1)+1$. If $p | a$, then $p \not |b$, so therefore $s_p(b)+s_p(a-b) = s_p(b-1)+s_p(a-b)+1 \ge s_p(a-1)+1$. So we're done. Therefore $m$ prime works. Now assume $m$ isn't prime. If $2 |n$ or $3|m$, use $n=\frac{m}{2}$ or $\frac{m}{3}$ and we reach a contradiction. Otherwise, $m \equiv 1,5$ (mod $6$). If $m$ is coprime with $6$, take $l$ an integer and $p$ a prime such that $m=(2l+1)(p)$. Take $n=m\left( \frac{l}{2l+1} \right)$. We can easily check $n$ lies on the required interval. Then, $m-2n = p$, and $n=lp$. It is very easy to check that $lp \not | \binom{lp}{p}$. Because we can reduce it to $p! | ((l-1)p+1)...(lp-1)$ and this is obviously false. Answer: $m$ prime. $\blacksquare$. Now, does anyone know necessary and sufficient conditions for $a | \binom{a}{b}$, given $1 \le b \le a-1$? Or is it an open problem?

thanks for your solution Juan Carlos Ortiz Rhoton

Lemma 1. We have $\frac{\gcd (m,n)}{n} \binom nm \in \mathbb{Z}$ for all positive integers $n \ge m$.

Lemma 2. We have $p^{k-1} \parallel \binom{p^k \cdot h}{p}$ for all prime $p$ and positive integer $h$.

Back to the problem. We consider two cases. Case 1. If $m$ is a prime then $\gcd (n,m-2n)=1$. Therefore by lemma 1 we have $n \mid \binom{n}{m-2n}$. Case 2. If $m$ is composite number. If $2|m$ then $m=2k$. Choose $n=k$ then $k|\binom{n}{m-2n}=\binom{k}{0}=1$ or $k=1$. Hence $m=2$, a contradiction. Thus, $2 \nmid m$. Let $m=pk$ with $k \ge 2, k \in \mathbb{Z}; p \ge 3, p \in \mathbb{P};2 \nmid pk$. Choose $n= \frac{p(k-1)}{2}$. We have $\binom{n}{m-2n}= \binom{ \frac{p(k-1)}{2}}{p}$ and $\frac{p(k-1)}{2} \mid \binom{ \frac{p(k-1)}{2}}{p}$. Applying lemma 2 we have $v_p \left( \binom{\frac{p(k-1)}{2}}{p} \right) = v_p\left( \frac{p(k-1)}{2} \right)-1$. Therefore, $m$ can not be a composite. Thus, $m$ is a prime number.

We can state that $m$ is a prime if and only if $n| \binom{n}{m-2n}$ for every $\frac m3 \le n \le \frac m2$. This is an old problem. I found this problem on 17 lectures about Fermat numbers. A generlization of this is: [Heiko Harborth,1977] Let $c \ge 2$ be any fixed integer. Suppose that $k>1$ is an integer that is not divisible by any prime number less than or equal to $c^2-c-1$. Then $k$ is a prime number if and only if $n| \binom{n}{k-cn}$ for all $n$ such that $\frac{k}{c+1} \le n \le \frac kc$.

We claim that the problem holds iff $m$ is prime. Suppose $m$ is even. Then taking $m=2k$, we get $k\mid\binom{k}{0}$ so $k=1$, and $m=2$. Now suppose $m$ is a multiple of three. Then let $m=3k$, so that $3k\mid\binom{k}{k}=1$ so $m=3$. Thus, we may assume $\gcd (m,6)=1$. Suppose $q<m$ is prime and $q\mid m$. Then take $n=\frac{m-q}{2}$. Then it follows $\frac{m-q}{2}\mid \binom{\frac{m-q}{2}}{q}$. But this is an obvious contradiction, as $v_q\left(\frac{\left(\frac{m-q}{2}\right)!}{\left(\frac{m-q}{2}-q\right)!}\right)=v_q\left(\frac{m-q}{2}\right)$, so $v_q\left(\binom{\frac{m-q}{2}}{q}\right)=v_q\left(\frac{m-q}{2}\right)-1$. Thus, the problem is reduced to $m$ prime. We claim all such $m$ work. Note that the problem claim is that $v_p\left(\binom{n}{m-2n}\right)\ge v_p(n)$ for all $p\mid n$ and $m/3<n<m/2$. Since $m$ is prime, if $p\mid n$, then $p\nmid m-2n$ and $p\nmid 3n-m$, as otherwise $p=m$ and $m\mid n$, a clear contradiction. Thus, $v_p((m-2n)!)=v_p((m-2n-1)!)$. Note that $v_p\left(\binom{n-1}{m-2n-1}\right)\ge 0$. On the other hand, $v_p(n)-v_p(m-2n)=v_p (n)$. Since $\binom{n}{m-2n}=\frac{n}{m-2n}\binom{n-1}{m-2n-1}$, we are done.

I claim the solution is all primes $m$. First, for $m$ even and composite take $n = \frac{m}{2}$. For $m$ divisible by $3$ and composite, take $n = \frac{m}{3}$. Else, if $m$ composite, we can write $m = p(2k+1)$. I claim that $n = pk$ does not divide $\dbinom{pk}{2pk+p-2pk} = \dbinom{pk}{p}$. To that end, just note that \[ v_p \left( \dbinom{kp}{p} \right) = \frac{s_p(p) + s_p(kp-p) - s_p(kp)}{p-1} \]We have $s_p(kp-p) \le s_p(kp) + s_p(-p) = s_p(kp) - s_p(p)$, thus $v_p \left( \dbinom{kp}{p} \right) = 0$ and the claim is proven. Now we will show all $m$ prime work. By a well-known binomial identity, $\dbinom{n-1}{m-2n-1} = \frac{m-2n}{n} \dbinom{n}{m-2n}$. If we show that $\gcd (m-2n, n) = 1$, then we would finish. If $m$ is even, then $m=2$ and the result is trivial. Else $m$ and $m-2n$ are odd and \[ \gcd(m-2n, n) = \gcd(2n, m-2n) = \gcd(m, n) = 1 \]as $m$ is prime and $n < m$, completing the proof.

We claim that all primes $p$ work. Assume $m$ is a prime $p > 2$, and consider $\frac {p - 2n}{n} \binom{n}{p-2n}$. Note this is just $\binom{n-1}{p-2n-1}$, which is an integer due to the restrictions ordained by the problem statement ($n \ge p - 2n$ and $p - 2n > 0$ since $p$ is odd). This means that $n$ divides $(p - 2n)\binom{n}{p-2n}$, but $n$ cannot divide $(p-2n)$ since $n$ does not divide $p$ (trivial case $n = 1$ omitted), so it follows that $n$ divides $\binom{n}{p-2n}$ for all $\frac{m}{3} \leq n \leq \frac{m}{2}$. When $p = 2$, $n$ must be $1$, so we are done. Now assume that $m$ is a composite number. We can actually set $n$ to be a multiple of $m - 2n$, in that if $n = k \cdot (m - 2n)$, then $n = \frac {k}{2k + 1} \cdot m$, provided that $m$ has an odd divisor. $\frac {k} {2k+1}$ takes a minimum at $\frac {1}{3}$ and is bounded above by $\frac{1}{2}$, so $n$ satisfies the conditions. Let $c = \frac {m} {2k+1}$. We can choose $k$ sufficiently so that $k$ is positive and $c$ is a prime divisor of $m$, since $m$ is composite. Then we want to prove that $kc$ does not divide $\binom{kc}{c} = k \cdot \binom{kc - 1}{c - 1}$. This is now equivalent to proving $c$ does not divide $\binom{kc - 1}{c - 1} = \frac{(kc - 1) \cdot (kc - 2) ... \cdot (kc - (c - 1))} {(c - 1) \cdot ... \cdot 1}$. But remember that $c$ is prime, and no multiple of $c$ appears in the numerator since $kc$ is a multiple but we start at $kc - 1$ but we don't reach $(k-1)c$. Thus $c$ cannot divide this binomial coefficient, and we are done with the case assuming $m$ has an odd prime divisor and is composite. The other case is when $m$ is a power of $2$ greater than $2$, but this is easy. Simply take $n = \frac {m} {2}$ and we are done. Thus, the only working $m$ are the primes.

Primes are obvious because you can easily write $\binom n{m-2n}=\frac n{3n-m}\binom{m-1}{m-2n}$ (thanks standard trick). For composite which are a multiple of $2$, just take $n=\frac m2$ and you get $n\mid 1$. For composite which are a multiple of $3$, just take $n=\frac m3$ and you get $n\mid 1$. Now, assume $p$ is the smallest prime factor of $m$ (which is at least 5). The idea is to make $n$ a multiple of $n-2m$, because it's easier to just go forward with Lagrange's. We let $q=\frac mp$, so then we essentially consider $q=2k+1$. The rest follows by consider $n=kp$. We show that $\binom {kp}p$ is not divisible by $p$. Note that if $p\nmid k$, then we are done as $p^2\nmid kp$. Furthermore, we have that if $p\mid k$, define $v=\nu_p(k)$. Then, we get that \[\nu_p((kp)!)-\nu_p(((k-1)p)!)-\nu_p(p!)=\nu_p(kp)-\nu_p(p)=\nu_p(k)\]which essentially shows that $pk\nmid\binom{kp}{k}$. Comments: This is very straightforward by small cases. In my opinion, it's not too bad of a problem. I think it's interesting and the fact that $p^{\nu_p(k)}\mid\mid \binom{kp}{p}$ is an interesting fact.

We claim the answer is all prime $m$. Note that $m = 2, m = 3$ work; suppose $m > 3$. Then, if $2 \mid m$, we require $\frac{m}{2} \mid \binom{\frac{m}{2}}{0}$, which is impossible for $m > 3$. Similarly, we cannot have $3 \mid m$ either. Suppose $m$ is composite. Note that $\left\lfloor \frac{m}{2} \right\rfloor - \left\lceil \frac{m}{3} \right\rceil \leq \frac{m}{2} - \frac{m}{3} - 2 = \frac{m}{6} - 2$. Let $p$ be the smallest prime dividing $m$, and let $k \nu_p(m)$. Since $\gcd(m, 6) = 1$, it follows that $\frac{m}{p^k} > 6$. This implies that $\frac{m}{6} - 2 > p^k$, meaning that we may find $n$ where $\frac{m}{3} \leq n \leq \frac{m}{2}$ such that $\nu_p(n) \geq k$. Therefore, for this $n$, we have $\nu_p(m - 2n) \geq \nu_p(n) = k$. Therefore, by Lucas's theorem, we have that $p \nmid \binom{n}{m - 2n}$, a contradiction. Therefore, we require $m$ to be prime. Now, suppose $m$ is prime. For any $\frac{m}{3} \leq n \leq \frac{m}{2}$, we require \begin{align*} n &\nmid \binom{n}{m - 2n}. \end{align*}Note that $\gcd(m - 2n, n) \leq \gcd(m - 2n + 2 \cdot n, n) = \gcd(m, n) = 1$. As $\binom{n}{m - 2n} = \frac{n}{m - 2n} \binom{n - 1}{m - 2n - 1}$. As $\gcd(m - 2n, n) = 1$ and $\binom{n}{m - 2n}$ is an integer, it follows that we must have $(m - 2n) \mid \binom{n - 1}{m - 2n - 1}$. Thus, $\binom{n}{m - 2n} = n \cdot \frac{\binom{n - 1}{m - 2n - 1}}{m - 2n}$, which is divisible by $n$. Thus, for all $\frac{m}{3} \leq n \leq \frac{m}{2}$, $n \mid \binom{n}{m - 2n}$, so we are done. $\Box$

Darn, I spent way too long on this problem. I thought it was going to be a difficult $v_p$ bash, but it turned out to be a lot simpler. So glad I finally solved this lol

Solution from Twitch Solves ISL (thought the answer was a finite set for the longest time!): The answer is $m$ prime. The canonical way to think of this is: Claim: The number $m$ works if and only if $\gcd(n,m) = 1$ for all $n \in [m/3,m/2]$. Proof. Fix $m$, and choose any prime $p$. The divisibility is rewritten as \[ \nu_p\left( (n-1)! \right) \ge \nu_p\left( (m-2n)! \right) + \left( \nu_p(2n)! \right) \]Define \[ f(n,q) = \left\lfloor \frac{n-1}{q} \right\rfloor - \left\lfloor \frac{m-2n}{q} \right\rfloor - \left\lfloor \frac{3n-m-1}{q} \right\rfloor. \]Then we have \[ \nu_p\left(\binom{n}{m-2n}\right) = \sum_{e \ge 1} f(n,p^e). \]We want this sum to be nonnegative for any prime $p$. The following properties are true: $f(n,q) \ge 0$ if $q \nmid n$, since then $\left\lfloor \frac{n-1}{q} \right\rfloor = \left\lfloor \frac nq \right\rfloor$. So there are no issues when $p \nmid n$. Now suppose $q$ is a prime power of a prime $p$ dividing $n$. We may explicitly compute \begin{align*} f(qk, q) &= (k-1)) - \left( \left\lfloor \frac{m-2(qk)}{q} \right\rfloor + \left\lfloor \frac{3(qk)-m-1}{q} \right\rfloor \right) \\ &= -1 - \left\lfloor \frac{m}{q} \right\rfloor - \left\lfloor \frac{-(m+1)}{q} \right\rfloor \\ &= \begin{cases} -1 & q \mid m \\ 0 & \text{otherwise}. \end{cases} \end{align*}In that case, it follows that $\nu_p(\tbinom{n}{m-2n})$ is the sum of some non-positive terms, at least one of which is $-1$ (namely the first term $f(n,p^1)=-1$.) So $m$ fails. In conclusion, $m$ works if and only $n$ never has a common prime factor with $m$, as desired. $\blacksquare$

We claim the answer is all primes. Note that it suffices to find all $m$ such that for all $n\in [m/3, m/2],$ $$N=\frac{(n-1)!}{(m-2n)!(3n-m)!} \in \mathbb Z \iff \sum_{i=1}^\infty \left \lfloor \frac{n-1}{p^i}\right\rfloor \ge \sum_{i=1}^\infty \left(\left\lfloor \frac{m-2n}{p^i} \right\rfloor + \left\lfloor \frac{3n-m}{p^i}\right\rfloor\right).$$Claim: $m$ and $n$ must be relatively prime. Proof. Suppose prime $p \mid m,n.$ Let $n=p^r \cdot x$ with $p\nmid x$ and $p^k\le x < p^{k+1}.$ Note then that for $1\le i\le r,$ $$\left\lfloor \frac{n-1}{p^i}\right\rfloor = \frac{n}{p^i} - 1 \qquad \left\lfloor \frac{m-2n}{p^i} \right\rfloor + \left\lfloor \frac{3n-m}{p^i}\right\rfloor \in\{n/p^i, n/p^i - 1\}.$$Where for $i=1$ the latter expression is precisely $n/p^i.$ As a result, for some positive integer $t,$ $$\sum_{i=r+1}^\infty \left \lfloor \frac{n-1}{p^i}\right\rfloor = \sum_{i=1}^k\left\lfloor \frac{x}{p^i}\right\rfloor - 1\ge t + \sum_{i=r+1}^\infty \left(\left\lfloor \frac{m-2n}{p^i} \right\rfloor + \left\lfloor \frac{3n-m}{p^i}\right\rfloor\right).$$Furthermore, $$\left\lfloor \frac{m-2n}{p^i} \right\rfloor + \left\lfloor \frac{3n-m}{p^i}\right\rfloor \ge \left\lfloor \frac{n}{p^i}\right\rfloor - 1 = \left\lfloor \frac{x}{p^{i-r}}\right\rfloor - 1,$$Implying $$\sum_{i=r+1}^\infty \left \lfloor \frac{n-1}{p^i}\right\rfloor = \sum_{i=1}^\infty \left\lfloor \frac{x}{p^i}\right\rfloor - 1\ge t + \sum_{i=r+1}^\infty \left(\left\lfloor \frac{m-2n}{p^i} \right\rfloor + \left\lfloor \frac{3n-m}{p^i}\right\rfloor\right)\ge t + \sum_{i=1}^k\left\lfloor \frac{x}{p^{i}}\right\rfloor - 1,$$Giving $t\le 0,$ contradiction. $\blacksquare$ Claim: $m$ must be prime. Proof. Note that clearly $2,3\nmid m$ since all $n\in[m/3, m/2]$ must be relatively prime to $m.$ Suppose that $m$ is composite and prime $q\mid m.$ Then, by virtue of gaps of size $q,$ we have $$q > \frac m2 - \frac m 3 = \frac m 6 \iff 6q > m,$$Implying that $m=5q$ and moreover $m < 30.$ It is easy to verify that for $m\in \{10,15,20,25\}$ that there exists such $n$ not relatively prime to $m,$ contradiction. $\blacksquare$ Claim: All prime $m$ suffices. Proof. Indeed, note that for all $p^i\nmid n,$ $$\left\lfloor \frac{m-2n}{p^i} \right\rfloor + \left\lfloor \frac{3n-m}{p^i}\right\rfloor \le \left\lfloor \frac{n}{p^i}\right\rfloor = \left\lfloor \frac{n-1}{p^i}\right\rfloor.$$Moreover, for $p^i\mid n,$ $p^i\nmid m-2n, 3n-m,$ implying $$\left\lfloor \frac{m-2n}{p^i} \right\rfloor + \left\lfloor \frac{3n-m}{p^i}\right\rfloor \le \frac{m-2n}{p^i} + \frac{3n-m}{p^i} - 1 = \frac{n}{p^i} - 1 = \left\lfloor \frac{n-1}{p^i}\right\rfloor.$$As a result, $N$ is an integer. $\blacksquare$

The answer is all prime numbers. Claim: No composite $m$ works. Proof: If $m$ is even then choose $n=\frac{m}{2}$ implying $\frac{m}{2}|1$ or $m=2$ which is a prime. For odd $m$ choose $n=\frac{m-p}{2}$ and by assumption,$n|\binom{n}{p}$ where $p$ is the smallest factor of $m$ hence $p|(n-1)(n-2)...........(n-p+1)$ which is impossible since $p|n$ Claim: Prime $m$ works. Proof: Clearly $p|(m-2n)\binom{n}{m-2n}$ and since $\gcd(m-2n,n)=1$,we are done.

Case 1. $m=2k$ for some positive integer $k>1$ We choose $n=k$ and found that \[ k\ \bigg| \ \binom{k}{0} \quad \text{which implies that} \ k=1 \implies m=2. \]Case 2. $m=2l+1$ for some positive integer $l$. Because $m \ge 3$ so that there is a prime number $p$ such that $p | m$. If $m>p$ since $2 \nmid m $ we can choose $\frac{m}{3}\le n=\frac{m-p}{2}\le \frac{m}{2}$ which makes \[n \ \bigg| \ \frac{n!}{p!\cdot (n-p)!} \implies p(p-1)\cdots(1)\ | \ (n-1)\cdots (n-p+1).\]Hence \[ p\ | \ (n-1)\cdots (n-p+1) \implies p \ \nmid \ n \implies p \ \nmid m-p \]which leads to contradiction since $p \ | \ m$. Therefore $m=p$. Next, we gonna proof that $m=p$ satisfies the condition for every odd prime $p$. Consider \[ n \ \bigg| \ \binom{n}{p-2n} \ \Leftrightarrow \ \nu_{q}{((p-2n)!)}+\nu_{q}{((3n-p)!)}\le\nu_{q}{((n-1)!)} \]If and only if \[\sum_{i=1}^{\infty}{\left\lfloor{\frac{n-1}{q^i}}\right\rfloor}\ge \sum_{i=1}^{\infty}{\left\lfloor{\frac{p-2n}{q^i}}\right\rfloor} +\sum_{i=1}^{\infty}{\left\lfloor{\frac{3n-p}{q^i}}\right\rfloor}\]for every prime $q$. Clearly, $p-2n,3n-p$ can't divide by $q$ together for every prime $q$ so that \begin{align*} \sum_{i=1}^{\infty}{\left\lfloor{\frac{n-1}{q^i}}\right\rfloor}&\ge\sum_{i=1}^{\infty}{\frac{n-1}{q^i}}=\sum_{i=1}^{\infty}{\left(\frac{p-2n}{q^i}+\frac{3n-p}{q^i}-\frac{1}{q^i}\right)} \\ &\ge \sum_{i=1}^{\infty}{\left\lfloor{\frac{p-2n}{q^i}}\right\rfloor} +\sum_{i=1}^{\infty}{\left\lfloor{\frac{3n-p}{q^i}}\right\rfloor} \end{align*}for every prime $p$. Therefore, all prime numbers are the solutions.

We claim only $m$ prime works. Notice that if $m = 2k$, or $m=3k$ taking $n=k$ $\implies$ $k \mid 1$. So $mdc(m,6) = 1$. Because $n$ $\mid$ ${n\choose m-2n}$ $=$ $\frac{n}{3n-m}$ ${n-1\choose m-2n}$ $(I)$ If $m$ is prime, because $mdc (3n-m,n)$ = $1$ we are done by $\frac {{n-1 \choose m-2n}}{3n-m}$ $\in$ $\mathbb{Z}$. Otherwise, if $m$ is composite, let $m$ = $p$ $\cdot$ $(2l+1)$, for some $p$ prime, and take $n$ = $pl$. Then, by (I): $V_{p}$ $({pl-1 \choose p})$ $\geq$ $V_{p}$ $(pl-p)$. So $V_{p}$ $({pl-1 \choose p})$ = $V_{p}$ $((pl-1)!) - V_{p}$ $(p!) - V_{p} ((pl-p-1)!)$ $=$ $V_{p}$ $((pl-p)!)$ - $V_{p} ((pl-p-1)!) - 1$ $\leq$ $V_{p}$ $((pl-p)!) - 1$, so $\frac {{n-1 \choose m-2n}}{3n-m}$ isn't an integer. Therefore $m$ must be prime as we wanted to show

The answer is all primes. Note that \[\binom{n}{m-2n}=\frac{n}{3n-m}\cdot \binom{n-1}{m-2n}\]Since $\gcd(n,3n-m)$, $n$ does divide that. Suppose $2\mid m$ and $m>2$ then clearly $n=m/2$ fails. Suppose $3\mid m$ and $m>3$ then $n=m/3$ clearly fails. If $m$ has a prime factor larger than $3$, let $m=(2a+1)k$ where $k$ is the smallest prime divisor then $n=ak$ gives \[\binom{ak}{k}=\frac{(ak)(ak-1)\dots (ak-k+1)}{k(k-1)\dots 1}\]Note that both the numerator and the denominator have $\nu_k$ of $1$ so we're done because $n=ak$ and $k\nmid \tbinom{ak}{k}$.

Precisely prime $m$. In the prime case, notice that $gcd(m - 2n, n) = 1$, so we are done by Kummer's theorem. Otherwise, notice if $2, 3 \mid m$ then choosing $n = \frac{m}{2}, \frac{m}{3}$ kills. Otherwise, suppose some $p \geq 5$ divides $m$, then simply choose $n = \frac{m-p}{2}$. Since $p \mid n$, notice also that $v_p {n \choose p} = v_p{\frac{n}{p} \choose 1} = v_p(n) - 1$, so the condition fails.