Find all positive integers $k$ such that for every $n\in\mathbb N$, if there are $k$ factors (not necessarily distinct) of $n$ so that the sum of their squares is $n$, then there are $k$ factors (not necessarily distinct) of $n$ so that their sum is exactly $n$.
1. **Verification for \( k = 1, 2, 3, 6 \):**

   - For \( k = 1 \):
     \[
     n = n
     \]
     This is trivially true since \( n \) is a factor of itself.

   - For \( k = 2 \):
     \[
     n = \left(\frac{n}{2}\right)^2 + \left(\frac{n}{2}\right)^2 = n
     \]
     This implies \( n \) must be even. If \( n \) is even, then \( n = \frac{n}{2} + \frac{n}{2} \) is valid.

   - For \( k = 3 \):
     \[
     n = \left(\frac{n}{3}\right)^2 + \left(\frac{n}{3}\right)^2 + \left(\frac{n}{3}\right)^2 = n
     \]
     This implies \( n \) must be divisible by 3. If \( n \) is divisible by 3, then \( n = \frac{n}{3} + \frac{n}{3} + \frac{n}{3} \) is valid.

   - For \( k = 6 \):
     \[
     n = \left(\frac{n}{6}\right)^2 + \left(\frac{n}{6}\right)^2 + \left(\frac{n}{6}\right)^2 + \left(\frac{n}{6}\right)^2 + \left(\frac{n}{6}\right)^2 + \left(\frac{n}{6}\right)^2 = n
     \]
     This implies \( n \) must be divisible by 6. If \( n \) is divisible by 6, then \( n = \frac{n}{6} + \frac{n}{6} + \frac{n}{6} + \frac{n}{6} + \frac{n}{6} + \frac{n}{6} \) is valid.

2. **Verification that no other \( k \) is possible:**

   - For \( k = 4 \):
     Consider \( n = (2 \cdot 7)^2 + (2 \cdot 7^2)^2 + 2 \cdot 7^2 \). This cannot be expressed as the sum of 4 positive divisors of \( n \).

   - For \( k = 5 \):
     Consider \( n = (7^2)^2 + 4 \cdot 7^2 \). This cannot be expressed as the sum of 5 positive divisors of \( n \).

   - For \( k > 6 \):
     Consider two cases:
     
     (1) \( k \) is not divisible by 4:
     Choose an odd \( a \in \mathbb{N} \) such that \( \min\{d \in \mathbb{N} \mid d \mid a^2(a^2 + k - 1), d > 6\} > 6k \) and \( 5 \nmid a^2(a^2 + k - 1) \). Then \( n = a^2(a^2 + k - 1) = (a^2)^2 + (k - 1)a^2 \) cannot be expressed as the sum of \( k \) positive divisors of \( n \).

     (2) \( k \) is divisible by 4:
     Choose an odd \( a \in \mathbb{N} \) such that \( \min\{d \in \mathbb{N} \mid d \mid a^2(4a^2 + k + 2), d > 6\} > 6k \) and \( 5 \nmid a^2(4a^2 + k + 2) \). Then \( n = a^2(4a^2 + k + 2) = (2a^2)^2 + (2a)^2 + (k - 2)a^2 \) cannot be expressed as the sum of \( k \) positive divisors of \( n \).

\(\blacksquare\)

The final answer is \( \boxed{ k = 1, 2, 3, 6 } \)