What is the smallest positive multiple of $225$ that can be written using
digits $0$ and $1$ only?
1. **Understanding the Problem:**
   We need to find the smallest positive multiple of \(225\) that can be written using only the digits \(0\) and \(1\).

2. **Factorization of 225:**
   \[
   225 = 15^2 = (3 \times 5)^2 = 3^2 \times 5^2
   \]
   Therefore, any multiple of \(225\) must be divisible by both \(9\) and \(25\).

3. **Divisibility Rules:**
   - A number is divisible by \(9\) if the sum of its digits is divisible by \(9\).
   - A number is divisible by \(25\) if its last two digits are \(00\), \(25\), \(50\), or \(75\). Since we can only use \(0\) and \(1\), the last two digits must be \(00\).

4. **Constructing the Number:**
   - The number must end in \(00\) to be divisible by \(25\).
   - The sum of the digits (excluding the trailing zeros) must be divisible by \(9\).

5. **Finding the Smallest Number:**
   - We start with the smallest number that ends in \(00\) and has a sum of digits divisible by \(9\).
   - The smallest such number is \(11111111100\) (nine \(1\)'s followed by two \(0\)'s).

6. **Verification:**
   - Check divisibility by \(9\):
     \[
     1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9 \quad (\text{which is divisible by } 9)
     \]
   - Check divisibility by \(25\):
     \[
     \text{The last two digits are } 00 \quad (\text{which is divisible by } 25)
     \]
   - Therefore, \(11111111100\) is divisible by both \(9\) and \(25\), and hence by \(225\).

Conclusion:
\[
\boxed{11111111100}
\]