cry. $ $ $ $

Not sure, but maybe it helps to note that $n=(k-1)!$ works for prime $k$. (This was motivated by ISL 2005 N4.)

Alright let's post up. With (i.e. ripped off of) flashsonic We claim all $k$ work. Fix $k$ in the sequel. Let $M = 10000^{1000+k!}$ be some really large number (make this bigger if necessary). Construction For any prime $p\leq k$: - If $p\mid k$, determine $b\geq 0$ and $1\leq c < p$ such that \[ cp^b\leq k-1 < (c+1)p^b \]We assert that $\boxed{n\equiv cp^b + kp^b \pmod{p^M}}$. - Otherwise, assert that $\boxed{p^M\mid n}$. Great. CRT and we've got ourselves an $n$. Fix it in the sequel. In addition, whenever we talk a prime $p$ from now on, if $p\mid k$, then the corresponding $b$ and $c$ are understood as defined above. Various observations to ensure our construction does what we want Fix any $p\leq k$. For all $1\leq r\leq p^{b+1}$: - If $p\mid k$ and $r = cp^b$ exactly, then $\nu_p(n-r) = \nu_p(kp^b) = \nu_p(r)+\nu_p(k)$. - Otherwise, $\nu_p(n-r) = \nu_p(r)$. Now, fix any $2\leq m\leq k$. \[ \nu_p\left(\binom{n}{m}\right) = \left(\sum_{1\leq r<m} \nu_p\left(\frac{n-r}{r}\right)\right) +\nu_p\left(\frac{n}{m}\right) \]If $p\mid k$ and $m>cp^b$, the summation is precisely $\nu_p(k)$, and \[ \nu_p\left(\binom{n}{m}\right) = \nu_p\left(\frac{kn}{m}\right) \]Otherwise, the summation vanishes, and \[ \nu_p\left(\binom{n}{m}\right) = \nu_p\left(\frac{n}{m}\right) \]In any case, if $n\mid \binom{n}{m}$, \[ \nu_p\left(\frac{kn}{m}\right)\geq \nu_p\left(\binom{n}{m}\right)\geq \nu_p(n) \]This works for all $p\leq k$, thus $m\mid k$. Conclusion This time, fix a prime $p\mid m$. - If $m\leq cp^b$, then $\nu_p\left(\binom{n}{m}\right)= \nu_p\left(\frac{n}{m}\right) < \nu_p(n)$, as desired. - Otherwise, $cp^b<m$ and $\nu_p\left(\frac{kn}{m}\right)\geq \nu_p(n)$ for all $p$. So $\frac{k}{m}$ is an integer. Yet, looking at $p\mid m$, $\frac{k}{m} < \frac{c+1}{c}\leq 2$. Thus in fact $m = k$, which works, as desired.

The answer is all $k$. We can easily check $k=p^t$ for $n\equiv p^{t-1} (\bmod\; p^{100t})$ Claim 1: $\nu_p((n-1)!) < \nu_p((n-m)!) + \nu_p(m!)$ if and only if $R(n, p^{\nu_p(m)+\lfloor \log_p(m-1) \rfloor}) =0 $ or $\ge m$. Proof: This inequality is equivalent to $\sum\limits_{j=n-m+1}^{n-1} \nu_p(j) \ge \sum\limits_{j=1}^{m-1} \nu_p(j) + \nu_p(m)$. Let $M=\max\limits_{j=n-m+1}^{n-1} \nu_p(j)$. All terms other than $M$ can be evenly paired to those in $ \sum\limits_{j=1}^{m-1} \nu_p(j)$ because they are at most $\lfloor \log_p(m-1) \rfloor$, except for one max (with value $\lfloor \log_p(m-1) \rfloor$) which is paired with $M$. Now we have $M$ on the LHS and $\nu_p(m) + \max_{j=1}^{m-1} \nu_p(j)=\nu_p(m)+\lfloor \log_p(m-1) \rfloor$ on the right hand side, as desired. Now, onto our construction. For our convenience, let $f_p(m)=p^{\nu_p(m)+\lfloor \log_p(m-1) \rfloor}$ We only look at $m$ st $rad(m)\mid rad(k)$ for now. First, if $f_p(m)>f_p(k)$, we can set $R(n,f_p(m))=R(n,f_p(m)-1) + p^{f_p(m)-1} \ge R(n,f_p(m)-1) + p^{f_p(k)} \ge k$, so we are good on those numbers. If for all $m<k$ and $rad(m)|rad(k)$, there exists $p\mid m$ such that $\lfloor \log_p(m) \rfloor < \lfloor \log_p(k) \rfloor$, then there exists $p^t$ such that $m<p^t<k$. In this case, set $R(n,f_p(k))=p^t$ (or $p^t-p$), which works. Otherwise there exist $m<k, rad(m)|rad(k), \nu_p(m)=\nu_p(k)$ for all primes $p$. This implies $m\mid k$, so $m\le \frac k2$. For each such $m$, we consider $R(n,f_p(k))=ap^t \in [\frac k2, k-1]$ for $a<p$. Note the $m$'s in this case are greater than the $m$'s in the previous case, so this works. (A rephrased, more cleaned-up version of the same argument: for $f_p(m)>f_p(k)$, we can set $R(n,f_p(m))=R(n,f_p(m)-1) + (p-1)f_p(m)/p \ge R(n,f_p(m)-1) + (p-1)f_p(k) \ge k$. Now we focus on $S=\{m|m<k, rad(m)|rad(k), f_p(m) \le f_p(k) \forall p\mid m\}$ We set $R(n,f_p(k))=ap^t$ where $t$ is equal to $ap^t$ where $t$ is the greatest number such that $p^t<k$ and $a$ is as large as possible.) Assume for contradiction that for some $m<k$ and $rad(m)|rad(k)$, for all $p\mid m$, $R(n,f_p(m)) \in [1,m-1]$. If $f_p(m)<k$ we are automatically done. Otherwise, $R(n,f_p(k))=R(n,f_p(m))$. We clearly have $\lfloor \log_p(m-1) \rfloor = \lfloor \log_p(k-1) \rfloor$, so $\nu_p(m)\le \nu_p(k)$ for all $p\mid m$. This implies that $m\mid k$, so $m\le \frac k2$, which means it is not a counterexample.) Now, for $rad(m)\nmid rad(k)$, pick $p\mid m, p\nmid k$ and force $p^{1000m} \mid n$. The conclusion follows by the Chinese Remainder Theorem. Wow my construction is similar to the above poster.

pad wrote: Find all positive integers $k > 1$ for which there exists a positive integer $n$ such that $\tbinom{n}{k}$ is divisible by $n$, and $\tbinom{n}{m}$ is not divisible by $n$ for $2\leq m < k$. Merlijn Staps The answer is all $\boxed{k \ge 2}$ and assume that $k >4$ since these can be verified manually;$(n,k)=(4,10)$ works. The proof proceeds in three parts from here:- $\textit{i)}$ All numbers co-prime with $k$. Let $\chi=\{a_1,..................,a_{\phi(k)=m} \}$ be all the numbers which are co-prime with $k$ We have further two cases:- $\textit{i.1)}$ $a_j$ is prime. Let $\mathcal{M}_j=\nu_3(p_j-1)$ then by CRT,we can make sure that $\nu_3(n)=1$;more specificaly the number of multiples of $3^z$ in $n!=\text{that in }{a_j!}$ then $$\nu_3(n-g) \ge \nu_3(-g) = \nu_3(g)$$so $$\nu_3(n!)-\nu_3 \left(\left(n-k \right)!\right)-\nu_3(k!) =\nu_3 \left( \frac{n(n-1)..........(n-(p-1))}{p(p-1)..........1} \right)=0$$so this works. $\textit{i.2)}$ If $a_j$ isn't prime. Then consider $$\nu_p(n!)-\nu_3 \left(\left(n-k \right)!\right)-\nu_p(k!) =\nu_p \left( \frac{n(n-1)..........(n-(p-1))}{p(p-1)..........1} \right)=0$$where $p^{m \le \nu_p(a_j)}|n$. $\textit{ii)}$ $m|k$ Then we proceed with a $v_2$+adding extra primes argument. Consider any random even number $x \equiv 2,4,6 \pmod 8 \in \{d_{\tau(k)-1},k \}$ and by CRT make sure that $$ n \equiv x \pmod {2^{\text{a very large number }}}$$Obviously this makes $k$ work but non of the divisors of $k$ work. $\textit{iii)}$ non-co-prime+non divisor:-This works directly from $\textit{ii)}$ The result follows from CRT from choosing $n$ such that \begin{align*}n \equiv 3,6 \pmod 9 \\n \equiv x \pmod {2^{\text{a very large number }}} \\ n \equiv 0 \pmod {p^{\nu_p(a_j)}} \end{align*},so we are done.$\blacksquare$

Solved with rama1728. All $\boxed{k \ge 2}$ work. For any prime $p \mid k,$ let $a$ be the greatest nonnegative integer where $rp^a \le k < (r+1)p^a$ for some positive integer $r$ where $p \nmid r.$ Then set $n \equiv (r+k)p^a$ modulo a very large power of $p.$ For any prime $q < k$ not dividing $k,$ let $n$ be a multiple of a very large power of $q.$ For any $2\le m\le k,$ note that $\tbinom{n}{m}$ is divisible by $n$ iff $$\sum_{i=1}^{m} \nu_p\left(\frac{n-i}{i}\right) \ge \nu_p\left(m \right)$$and $$\sum_{i=1}^{m} \nu_q\left(\frac{n-i}{i}\right) \ge \nu_q\left(m \right)$$for all primes $p \mid k$ and all primes $q < k$ not dividing $k.$ For all $q,$ we defined $n$ where $\nu_q(i) = \nu_q(n-i)$ for all $1 \le i \le k.$ So no $q$ can divide $m.$ For all $p,$ we defined $n$ where $\nu_p(i) = \nu_p(n-i)$ for all $1 \le i \le k$ unless $i = rp^a,$ in which case $ \nu_p(n-rp^a) - \nu_p(rp^a) = \nu_p(k).$ So we require $\nu_p(k) \ge \nu_p(m)$ and $m \ge rp^a$ for all $p \mid m.$ This can only happen if $m$ is a divisor of $k,$ but the only divisor of $k$ satisfying the size conditions is $k$ itself, which works. $\square$

Solved on plane. All $\boxed{k \ge 2}$ works. Construction: For all $p \leq k$. If $p \mid k$, then let $1 \leq r \leq p-1$ so that $rp^a \leq k < (r+1)p^e$, then set $n \equiv (k+r)p^e \pmod p^M$ where $M$ is very large. If $p \not\mid k$ then $p^M \mid n$ for some very large $M$ Proof that construction works: Notice that $n \mid \binom{n}{m}$ iff $\sum_{1}^{m} \nu_p\left(\frac{n-i}{i}\right) \geq \nu_p(m)$ for all primes, now if $p \not\mid k$, then $\nu_q(i) = \nu_q(n-i)$ for all $1 \le i \le k.$ so $q \not\mid m$, For all $p \mid k$ we have $\nu_p(i) = \nu_p(n-i)$ for all $1 \le i \le k$ unless $i = rp^e,$, where in we have $\nu_p(n-rp^e)=\nu_p(k)+\nu_p(rp^e)$. So we must have $\nu_p(k) \geq \nu_p(m)$ and $m \geq rp^e$ for all $p \mid m$, which can only happen if it is divisor of $k$, but due to size restrictions it must be $k$, as desired!