Consider the binomial coefficients $\binom{n}{k}=\frac{n!}{k!(n-k)!}\ (k=1,2,\ldots n-1)$. Determine all positive integers $n$ for which $\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}$ are all even numbers.
1. **Understanding the Problem:**
   We need to determine all positive integers \( n \) for which the binomial coefficients \(\binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n-1}\) are all even numbers.

2. **Analyzing the Binomial Coefficients:**
   The binomial coefficient \(\binom{n}{k}\) is given by:
   \[
   \binom{n}{k} = \frac{n!}{k!(n-k)!}
   \]
   We need to find \( n \) such that all these coefficients are even.

3. **Using Lucas' Theorem:**
   Lucas' theorem states that for non-negative integers \( n \) and \( k \) and a prime \( p \), the binomial coefficient \(\binom{n}{k}\) modulo \( p \) can be determined by the base \( p \) representations of \( n \) and \( k \). Specifically, if \( n = n_0 + n_1 p + n_2 p^2 + \cdots \) and \( k = k_0 + k_1 p + k_2 p^2 + \cdots \), then:
   \[
   \binom{n}{k} \equiv \prod_{i} \binom{n_i}{k_i} \pmod{p}
   \]
   For \( p = 2 \), this simplifies to:
   \[
   \binom{n}{k} \equiv \prod_{i} \binom{n_i}{k_i} \pmod{2}
   \]
   where \( n_i \) and \( k_i \) are the binary digits of \( n \) and \( k \), respectively.

4. **Binary Representation and Powers of Two:**
   If \( n \) is a power of two, say \( n = 2^k \), then the binary representation of \( n \) is \( 100\ldots0 \) (a 1 followed by \( k \) zeros). For any \( 1 \leq k \leq n-1 \), the binary representation of \( k \) will have at least one 1 in a position where \( n \) has a 0. Thus, \(\binom{n}{k}\) will be even because at least one of the \(\binom{n_i}{k_i}\) will be \(\binom{0}{1} = 0\).

5. **Non-Power of Two:**
   Suppose \( n \) is not a power of two. Then \( n \) has more than one 1 in its binary representation. Let the leftmost 1 be at position \( j \) and another 1 at position \( i \). Construct \( m \) such that in binary, \( m \) has a 1 at position \( i \) and 0s elsewhere. According to Lucas' theorem, \(\binom{n}{m}\) will be odd because \(\binom{1}{1} = 1\) and \(\binom{0}{0} = 1\).

6. **Conclusion:**
   Therefore, \( n \) must be a power of two for all binomial coefficients \(\binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n-1}\) to be even.

\(\blacksquare\)

The final answer is \( \boxed{ n \in \{2^k \mid k \in \mathbb{N}\} } \).