Find all the real numbers $ N$ that satisfy these requirements:

1. Only two of the digits of $ N$ are distinct from $ 0$, and one of them is $ 3$.

2. $ N$ is a perfect square.
1. **Identify the form of \( N \)**:
   - \( N \) must have only two digits distinct from 0, one of which is 3.
   - \( N \) is a perfect square.

2. **Consider the simplest case without zeros**:
   - If there are no zeros, the only possible \( N \) is 36, since 36 is a perfect square and contains the digit 3.

3. **Consider cases with zeros**:
   - If \( N \) has zeros, it can be written in the form \( 3000\ldots0a \) or \( a000\ldots003 \), where \( a \) is a digit from 1 to 9.

4. **Check modulo conditions**:
   - For \( N \) to be a perfect square, it must satisfy certain modular conditions.
   - The square residues modulo 9 are 0, 1, 4, and 7. Therefore, \( 3 + a \) must be one of these residues.
   - Possible values for \( a \) are 6, 7, 1, and 4.

5. **Check modulo 4**:
   - For \( a000\ldots003 \), the number is congruent to 3 modulo 4, which is not a square residue modulo 4.
   - Therefore, \( a000\ldots003 \) cannot be a perfect square.

6. **Check possible candidates**:
   - The only possible candidates are \( 3000\ldots01 \), \( 3000\ldots04 \), \( 3000\ldots06 \), and \( 3000\ldots07 \).
   - However, the numbers ending in 6 and 7 do not work modulo 4, so only the first two are possible candidates.

7. **Verify the candidates**:
   - For \( 3000\ldots01 \) and \( 3000\ldots04 \), we need to check if they can be written as \( x^2 \).
   - \( 3 \times 10^n = x^2 - 1 = (x - 1)(x + 1) \).
   - This requires checking the greatest common divisor (gcd) conditions, which can be tedious.

8. **Conclude the solution**:
   - The only integers that satisfy the conditions are \( 36 \times 100^n \), where \( n \) is any integer \( \ge 0 \).

The final answer is \( \boxed{ 36 \times 100^n } \), where \( n \) is any integer \( \ge 0 \).