Given an integer $n>2$ and an integer $a$, if there exists an integer $d$ such that $n\mid a^d-1$ and $n\nmid a^{d-1}+\cdots+1$, we say [i]$a$ is $n-$separating[/i]. Given any n>2, let the [i]defect of $n$[/i] be defined as the number of integers $a$ such that $0<a<n$, $(a,n)=1$, and $a$ is not [i] $n-$separating[/i]. Determine all integers $n>2$ whose defect is equal to the smallest possible value. 
1. **Prime Case:**
   - Let \( n = p \) be a prime.
   - We need to show that \( T_n = \{1\} \).
   - If \( a = 1 \), we can take \( d \) such that \( p \nmid d \).
   - If \( a \neq 1 \), then \( p \mid a^d - 1 = (a-1)(a^{d-1} + \cdots + 1) \).
   - Since \( a < p \), \( p \mid a^{d-1} + \cdots + 1 \), hence \( a \notin S_n \).

2. **Prime Powers Case:**
   - Let \( n = p^\ell \) for some prime \( p > 2 \).
   - If \( a \in T_n \), then \( p \mid a - 1 \).
   - Using the Lifting the Exponent Lemma:
     \[
     v_p(a^d - 1) = v_p(a - 1) + v_p(d)
     \]
   - Select \( d = p^{\ell - v_p(a - 1)} \), then \( p^\ell \mid\mid a^d - 1 \) and \( p^\ell \nmid a^{d-1} + \cdots + 1 \).
   - Hence, \( S_n = T_n \).

3. **Composite Numbers with Coprime Conditions:**
   - Let \( n = p_1^{\alpha_1} \cdots p_k^{\alpha_k} \) where \( (p_i, p_j - 1) = 1 \) for all \( i, j \).
   - For \( a \in T_n \), define \( I = \{i : p_i \mid a - 1\} \).
   - Using the Lifting the Exponent Lemma:
     \[
     v_{p_i}(a^d - 1) = v_{p_i}(a - 1) + v_{p_i}(d)
     \]
   - Select \( v_{p_i}(d) = \alpha_i - v_{p_i}(a - 1) \) if \( v_{p_i}(a - 1) < \alpha_i \), otherwise ensure \( v_{p_i}(d) < \alpha_i \).
   - Use the Chinese Remainder Theorem to find \( d \) such that:
     \[
     d \equiv p_i^{\beta_i} \pmod{p_i^{\beta_i + 1}} \quad \forall i \in I \quad \text{and} \quad d \equiv 0 \pmod{\varphi(p_j^{\alpha_j})}
     \]
   - Thus, \( a \in T_n \Rightarrow a \in S_n \).

4. **Composite Numbers with Non-Coprime Conditions:**
   - Let \( n = p_1^{\alpha_1} \cdots p_k^{\alpha_k} \) with \( p_i^{\alpha_i} \nmid p_j - 1 \) for all \( i, j \).
   - For \( a \in T_n \), define \( I = \{i : p_i \mid a - 1\} \).
   - Set \( p_i^{\alpha_i - 1} \mid\mid d \) for all \( i \in I \) and \( \varphi(p_j^{\alpha_j}) \mid d \) for all \( j \).
   - Use the Chinese Remainder Theorem to find \( d \) such that:
     \[
     d \equiv p_i^{\beta_i} \pmod{p_i^{\beta_i + 1}} \quad \forall i \in I \quad \text{and} \quad d \equiv 0 \pmod{\varphi(p_j^{\alpha_j})}
     \]
   - Thus, \( a \in T_n \Rightarrow a \in S_n \).

5. **Composite Numbers with Divisibility Conditions:**
   - Let \( n = p_1^{\alpha_1} \cdots p_k^{\alpha_k} \) with \( p_i^{\alpha_i} \mid p_j - 1 \) for some \( i, j \).
   - Let \( g_j \) be primitive roots modulo \( p_j^{\alpha_j} \) for all \( j \neq i \).
   - Take \( a \in T_n \) such that:
     \[
     a \equiv 1 \pmod{p_i^{\alpha_i}} \quad \text{and} \quad a \equiv g_j \pmod{p_j^{\alpha_j}} \quad \forall j \neq i
     \]
   - For any \( d \) with \( n \mid a^d - 1 \), we have \( p_j - 1 \mid d \).
   - Since \( v_{p_i}(a^{d-1} + \cdots + 1) = v_{p_i}(d) \geq \alpha_i \), we get \( p_i^{\alpha_i} \mid a^{d-1} + \cdots + 1 \).
   - Thus, \( n \mid a^d - 1 \Rightarrow n \mid a^{d-1} + \cdots + 1 \), so \( a \notin S_n \).

The final answer is \( \boxed{ n } \) such that \( n \) is a prime or a prime power.