For each positive integer $k$, let $d(k)$ be the number of positive divisors of $k$ and $\sigma(k)$ be the sum of positive divisors of $k$. Let $\mathbb N$ be the set of all positive integers. Find all functions $f: \mathbb{N} \to \mathbb N$ such that \begin{align*}
			f(d(n+1)) &= d(f(n)+1)\quad \text{and} \\
			f(\sigma(n+1)) &= \sigma(f(n)+1)
		\end{align*}
	for all positive integers $n$.
1. **Lemma 1:** We need to show that \( x = d(x+1) \) if and only if \( x \in \{2, 3\} \).

   **Proof:** Let \( x+1 = 2^a \cdot 3^b \cdot p_1^{e_1} \cdots p_k^{e_k} \), where \( p_i \) are primes greater than 3. Then, the condition \( x = d(x+1) \) translates to:
   \[
   2^a \cdot 3^b \cdot p_1^{e_1} \cdots p_k^{e_k} = (a+1)(b+1)(e_1+1) \cdots (e_k+1) + 1
   \]
   Note that for any prime \( p \geq 5 \), we have \( p^k \geq 4k+1 \), \( 3^a \geq 2a+1 \), and \( 2^a \geq \max\{2a, a+1\} \). If \( x+1 \) has a prime divisor \( \geq 5 \) or \( e_1 \geq 1 \), then:
   \[
   x+1 \geq (a+1)(2b+1)(4e_1+1) \cdots (4e_k+1)
   \]
   This implies:
   \[
   x+1 \geq (a+1)(b+1)(e_1+1) \cdots (e_k+1) + 3e_1 > d(x+1) + 1
   \]
   which is a contradiction. Therefore, \( x+1 = 2^a \cdot 3^b \). We then have:
   \[
   (a+1)(b+1) + 1 = 2^a \cdot 3^b
   \]
   This implies:
   \[
   (a+1)(b+1) + 1 \geq 2a(2b+1) + 1
   \]
   which simplifies to:
   \[
   2 \geq 3ab + a
   \]
   Thus, the only solutions are \( \{a, b\} = \{2, 0\} \) or \( \{0, 1\} \). Therefore, \( x \in \{2, 3\} \).

2. Given the conditions:
   \[
   f(d(n+1)) = d(f(n)+1) \quad \text{and} \quad f(\sigma(n+1)) = \sigma(f(n)+1)
   \]
   Let \( n = 3 \) in the first condition. From Lemma 1, we get \( f(3) \in \{2, 3\} \).

3. Let \( n = 1 \) in the second condition. We get:
   \[
   \sigma(f(1)+1) = f(3) \in \{2, 3\}
   \]
   Since there is no positive \( x \) such that \( \sigma(x) = 2 \), we must have \( f(3) = 3 \). Therefore:
   \[
   \sigma(f(1)+1) = 3 \implies f(1) = 1
   \]

4. **Claim 1:** \( f(2^k-1) = 2^k-1 \) for all \( k \).

   **Proof:** We will prove this by induction. The base cases \( n = 1, 2 \) are already proved. Assume \( f(2^k-1) = 2^k-1 \). Let \( n \rightarrow 2^k-1 \) in the second condition. We get:
   \[
   f(\sigma(2^k)) = \sigma(f(2^k-1)+1) = \sigma(2^k) \implies f(2^{k+1}-1) = 2^{k+1}-1
   \]
   This completes the induction.

5. For any \( k \), let \( x \rightarrow 2^k -1 \) in the first condition. We get:
   \[
   f(d(2^k)) = d(f(2^k-1)+1) = d(2^k) \implies f(k+1) = k+1
   \]
   Thus:
   \[
   f(n) = n \quad \forall n
   \]

The final answer is \( \boxed{ f(n) = n } \) for all \( n \).