If the representation of  a positive number as a product of powers of distinct prime numbers contains no even powers other than $0$s, we will call the number singular. At most how many consequtive singular numbers are there?

$ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ \text{None}$
1. **Understanding the definition of singular numbers**:
   A number is singular if its prime factorization contains no even powers other than 0. This means that in the prime factorization of a singular number, each prime factor must appear to an odd power.

2. **Analyzing the problem**:
   We need to determine the maximum number of consecutive singular numbers. To do this, we need to consider the properties of numbers and their prime factorizations.

3. **Considering the properties of consecutive numbers**:
   Out of any eight consecutive positive integers, at least one of them is divisible by 4 but not by 8. This is because every fourth number is divisible by 4, and among any eight consecutive numbers, there will be at least one such number. 

4. **Checking the prime factorization of numbers divisible by 4 but not by 8**:
   If a number is divisible by 4 but not by 8, it can be written in the form \(4k\) where \(k\) is an odd number. The prime factorization of \(4k\) is \(2^2 \cdot k\). Since \(2^2\) is an even power, such a number cannot be singular.

5. **Finding a sequence of consecutive singular numbers**:
   We need to find the longest sequence of consecutive numbers that do not include a number of the form \(4k\) where \(k\) is odd. 

6. **Example of a sequence of seven consecutive singular numbers**:
   Consider the sequence \(37, 38, 39, 40, 41, 42, 43\):
   - \(37\) is prime.
   - \(38 = 2 \cdot 19\) (both primes to the power of 1).
   - \(39 = 3 \cdot 13\) (both primes to the power of 1).
   - \(40 = 2^3 \cdot 5\) (2 to the power of 3, which is odd, and 5 to the power of 1).
   - \(41\) is prime.
   - \(42 = 2 \cdot 3 \cdot 7\) (all primes to the power of 1).
   - \(43\) is prime.

   This sequence contains seven consecutive singular numbers.

7. **Verifying that eight consecutive singular numbers are not possible**:
   Since any sequence of eight consecutive numbers must include a number of the form \(4k\) where \(k\) is odd, and such a number is not singular, it is impossible to have eight consecutive singular numbers.

Conclusion:
The maximum number of consecutive singular numbers is 7.

The final answer is \(\boxed{7}\)