Vukasin,  Dimitrije, Dusan, Stefan and Filip asked their teacher to guess three consecutive positive integers, after these true statements:
Vukasin: " The sum of the digits of one number is prime number. The sum of the digits of another of the other two is, an even perfect number.($n$ is perfect if $\sigma\left(n\right)=2n$). The sum of the digits of the third number equals to the number of it's positive divisors".
Dimitrije:"Everyone of those three numbers has at most two digits equal to $1$ in their decimal representation".
Dusan:"If we add $11$ to exactly one of them, then we have a perfect square of an integer"
Stefan:"Everyone of them has exactly one prime divisor less than $10$".
Filip:"The three numbers are square free".
Professor found the right answer. Which numbers did he mention?
We have given the following conditions about three positive integers:

(C1) The numbers are consecutive.
(C2) The sum of the digits of one number is a prime.
(C3) The sum of the digits of another number is an even perfect number.
(C4) The sum of the digits of the third number equals the number of its positive divisors.
(C5) None of the three numbers has more than two digits equal to 1.
(C6) If 11 is added to one of the numbers, the sum is a perfect square.
(C7) The three numbers have exactly one prime divisor less than 10.
(C8) The three numbers are square-free.

We need to find the three consecutive numbers that satisfy all these conditions.

1. **Identify the smallest number \( N \):**
   Let \( N \) be the smallest of the three numbers. Then the numbers are \( N, N+1, N+2 \).

2. **Sum of digits conditions:**
   - The sum of the digits of one number is a prime.
   - The sum of the digits of another number is an even perfect number.
   - The sum of the digits of the third number equals the number of its positive divisors.

3. **Square-free condition:**
   Since the numbers are square-free, they do not have any repeated prime factors.

4. **Prime divisor condition:**
   Each number has exactly one prime divisor less than 10. The primes less than 10 are 2, 3, 5, and 7.

5. **Perfect square condition:**
   If 11 is added to one of the numbers, the result is a perfect square.

6. **Sum of digits analysis:**
   - The sum of the digits of one number is a prime.
   - The sum of the digits of another number is an even perfect number. The only even perfect number less than 10 is 6 (since 6 = \(2^1(2^2 - 1)\)).
   - The sum of the digits of the third number equals the number of its positive divisors. For a number to have a sum of digits equal to the number of its divisors, it must be a power of 2 (since the number of divisors of \(2^k\) is \(k+1\)).

7. **Check possible sums of digits:**
   - The sums of the digits of the three numbers must be consecutive integers.
   - The possible sums of the digits are 4, 5, 6 or 6, 7, 8.

8. **Perfect square condition:**
   - If \( N + 11 \) is a perfect square, then \( N + 11 = x^2 \) for some integer \( x \).
   - This implies \( N = x^2 - 11 \).

9. **Check the last digits:**
   - The last digits of the numbers cannot be 0 (since they are square-free and cannot have both 2 and 5 as prime factors).

10. **Prime factorization:**
    - The numbers must have exactly one prime divisor less than 10.

11. **Find the numbers:**
    - By checking the conditions, we find that the numbers 2013, 2014, and 2015 satisfy all the conditions:
      - 2013: Sum of digits = 6 (even perfect number).
      - 2014: Sum of digits = 7 (prime number).
      - 2015: Sum of digits = 8 (number of divisors = 8).

12. **Verify the conditions:**
    - 2013, 2014, and 2015 are consecutive.
    - The sum of the digits of 2013 is 6 (even perfect number).
    - The sum of the digits of 2014 is 7 (prime number).
    - The sum of the digits of 2015 is 8 (number of divisors = 8).
    - None of the numbers have more than two digits equal to 1.
    - Adding 11 to 2014 gives 2025, which is \(45^2\) (a perfect square).
    - Each number has exactly one prime divisor less than 10:
      - 2013: Prime factors are 3, 11, 61.
      - 2014: Prime factors are 2, 19, 53.
      - 2015: Prime factors are 5, 13, 31.
    - All numbers are square-free.

The final answer is \(\boxed{2013, 2014, 2015}\).