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 $2\leq m < k$.

[i]Merlijn Staps[/i]
1. **Understanding the Problem:**
   We need to find all positive integers \( k > 1 \) for which there exists a positive integer \( n \) such that \( \binom{n}{k} \) is divisible by \( n \), and \( \binom{n}{m} \) is not divisible by \( n \) for \( 2 \leq m < k \).

2. **Divisibility Condition:**
   Recall that \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). For \( \binom{n}{k} \) to be divisible by \( n \), \( n \) must divide the numerator \( n! \) but not the denominator \( k!(n-k)! \).

3. **Prime Factorization Insight:**
   Let \( p \) be a prime dividing \( k \). We need to consider the highest power of \( p \) dividing \( n \), denoted \( \nu_p(n) \). We will use the properties of \( \nu_p \) (the p-adic valuation) to analyze the divisibility conditions.

4. **Setting \( n \):**
   Let \( a \) be the greatest nonnegative integer such that \( rp^a \leq k < (r+1)p^a \) for some positive integer \( r \) where \( p \nmid r \). Set \( n \equiv (r+k)p^a \) modulo a very large power of \( p \).

5. **Analyzing \( \binom{n}{m} \):**
   For any prime \( q < k \) not dividing \( k \), let \( n \) be a multiple of a very large power of \( q \). For any \( 2 \leq m \leq k \), note that \( \binom{n}{m} \) is divisible by \( n \) if and only if:
   \[
   \sum_{i=1}^{m} \nu_p\left(\frac{n-i}{i}\right) \ge \nu_p(m)
   \]
   and
   \[
   \sum_{i=1}^{m} \nu_q\left(\frac{n-i}{i}\right) \ge \nu_q(m)
   \]
   for all primes \( p \mid k \) and all primes \( q < k \) not dividing \( k \).

6. **Ensuring Non-Divisibility for \( m < k \):**
   For all \( q \), we defined \( n \) where \( \nu_q(i) = \nu_q(n-i) \) for all \( 1 \leq i \leq k \). Thus, no \( q \) can divide \( m \).

7. **Ensuring Divisibility for \( k \):**
   For all \( p \), we defined \( n \) where \( \nu_p(i) = \nu_p(n-i) \) for all \( 1 \leq i \leq k \) 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.

8. **Conclusion:**
   Therefore, all \( k \ge 2 \) work.

The final answer is \( \boxed{ k \ge 2 } \).