Find all positive integers $a$ so that for any $\left \lfloor \frac{a+1}{2} \right \rfloor$-digit number that is composed of only digits $0$ and $2$ (where $0$ cannot be the first digit) is not a multiple of $a$. 

To solve the problem, we need to find all positive integers \( a \) such that any \(\left \lfloor \frac{a+1}{2} \right \rfloor\)-digit number composed only of the digits \(0\) and \(2\) (where \(0\) cannot be the first digit) is not a multiple of \( a \).

1. **Initial Analysis**:
   - Let \( S \) be the set of all positive integers satisfying the problem.
   - We need to ensure that for any \(\left \lfloor \frac{a+1}{2} \right \rfloor\)-digit number composed of \(0\) and \(2\), the number is not divisible by \( a \).

2. **Claim 1**: If \( a = 2^p \cdot 5^q \cdot s \in S \), then \( s \in S \).
   - If \( s = 1 \) and \((p, q) \neq (1, 0)\), then \(\left \lfloor \frac{a+1}{2} \right \rfloor > \max \{p, q\}\).
   - Consider the number \(\overline{2\ldots20\ldots0}\) (with \(\max(p, q)\) zeros). This number is divisible by \( a \), leading to a contradiction.
   - If \((p, q) = (1, 0)\), then \(2 \mid a = 2\), which is a contradiction.

3. **Claim 2**: \( s \) has at most 2 prime divisors.
   - Assume \( s \) has at least 3 prime divisors.
   - **Case 1**: If \( 3 \mid s \), then \(\frac{10^{3 \cdot \frac{\varphi(s)}{4}} - 1}{9} \mid s\).
     - Using \( v_3 \) valuation, we get \( v_3\left(\frac{10^{3 \cdot \frac{\varphi(s)}{4}} - 1}{9}\right) \geq v_3(s) \).
   - **Case 2**: If \((s, 3) = 1\), then \(\frac{10^{\frac{\varphi(s)}{4}} - 1}{9} \mid s\).
     - For any \( p \neq 3 \), \( p \mid s \Rightarrow 10^{\frac{\varphi(s)}{4}} - 1 \mid p^{v_p(s)} \).

4. **Claim 3**: \( s = 3^r \cdot p^s \) (where \( p > 5 \) and \( p \) is a prime).
   - Assume \( s = p^s \cdot q^t \) (where \( p, q > 3 \) and \( p, q \) are primes).
   - \(\frac{10^{\frac{\varphi(s)}{2}} - 1}{9} \mid s\).

5. **Claim 4**: \( p^s \notin S \) (where \( p > 3 \) is a prime).
   - \( 2 \cdot \frac{10^{\frac{p-1}{2}} - 1}{9} \mid p \) or \( 2 \cdot \left[10^{\frac{p+1}{2}} + 1\right] \mid p \).

6. **Claim 5**: \( s = 3^t \cdot p^s \) (where \( s, t \geq 1 \)) is not in \( S \).
   - **Case 1**: \( 10^{\frac{p-1}{2}} + 1 \mid p \Rightarrow 10^{p^{s-1} \cdot \frac{p-1}{2}} + 1 \mid p^s \).
   - **Case 2**: \( 10^{\frac{p-1}{2}} - 1 \mid p \Rightarrow 10^{p^{s-1} \cdot \frac{p-1}{2}} - 1 \mid p^s \).

7. **Claim 6**: \( s = 3^t \in S \) then \( t < 3 \).
   - By induction on \( t > 2 \), we prove that there exists a number \( u_t \) with at most \(\frac{3^3 - 1}{2} \cdot 3^{t-3}\) digits such that \( u_t \mid t \).

8. **Verification**:
   - Check that \( 3 \) and \( 9 \) satisfy the problem.
   - If \( a = 2^p \cdot 5^q \cdot 3 \) or \( 2^p \cdot 5^q \cdot 9 \), but if \( p+q > 0 \), we have a contradiction as shown in Claim 1.

The final answer is \( \boxed{ 3, 9 } \).