Determine all positive integers $n > 2$ , such that \[ \frac{1}{2} \varphi(n) \equiv 1 ( \bmod 6)  \]
To determine all positive integers \( n > 2 \) such that 
\[ \frac{1}{2} \varphi(n) \equiv 1 \pmod{6}, \]
we need to analyze the properties of the Euler's totient function \(\varphi(n)\) and its behavior under different forms of \( n \).

1. **Understanding the Euler's Totient Function**:
   The Euler's totient function \(\varphi(n)\) counts the number of integers up to \( n \) that are coprime with \( n \). For a prime \( p \) and integer \( k \), we have:
   \[
   \varphi(p^k) = p^k - p^{k-1} = p^{k-1}(p-1).
   \]
   For \( n = p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m} \), where \( p_i \) are distinct primes, we have:
   \[
   \varphi(n) = n \left(1 - \frac{1}{p_1}\right) \left(1 - \frac{1}{p_2}\right) \cdots \left(1 - \frac{1}{p_m}\right).
   \]

2. **Analyzing the Congruence**:
   We need:
   \[
   \frac{1}{2} \varphi(n) \equiv 1 \pmod{6}.
   \]
   This implies:
   \[
   \varphi(n) \equiv 2 \pmod{12}.
   \]

3. **Case Analysis**:
   - **Case 1: \( n = 4k \)**:
     If \( n = 4k \), then \( \varphi(n) = \varphi(4k) = 2\varphi(2k) \). For \( \varphi(n) \equiv 2 \pmod{12} \), we need \( \varphi(2k) \equiv 1 \pmod{6} \). This is possible if \( 2k = 2 \), giving \( n = 4 \).

   - **Case 2: \( n = p^k \) or \( n = 2p^k \)**:
     For \( n = p^k \), we have \( \varphi(n) = p^{k-1}(p-1) \). For \( n = 2p^k \), we have \( \varphi(n) = 2p^{k-1}(p-1) \). We need:
     \[
     p^{k-1}(p-1) \equiv 2 \pmod{12}.
     \]
     This implies \( p-1 \equiv 2 \pmod{12} \), so \( p \equiv 3 \pmod{12} \) or \( p \equiv 5 \pmod{12} \).

4. **Specific Solutions**:
   - For \( p = 3 \), \( k = 1 \), we get \( n = 3 \).
   - For \( p \equiv 2 \pmod{3} \) and \( k \) even, we get \( n = p^{2k} \) or \( n = 2p^{2k} \).

5. **Combining Results**:
   The solutions are:
   \[
   n = 3, 4, 6, p^{2k}, 2p^{2k} \text{ for } k \equiv 11 \pmod{12}.
   \]

The final answer is \(\boxed{n = 3, 4, 6, p^{2k}, 2p^{2k} \text{ for } k \equiv 11 \pmod{12}}\).