Find all increasing sequences $a_1,a_2,a_3,...$ of natural numbers such that for each $i,j\in \mathbb N$, number of the divisors of $i+j$ and $a_i+a_j$ is equal. (an increasing sequence is a sequence that if $i\le j$, then $a_i\le a_j$.)
1. **Claim 1:** If \( p \) is a prime, then \( a_{2^{p-2}} = 2^{p-2} \).

   **Proof:** From \( P(2^{p-2}, 2^{p-2}) \), we have:
   \[
   \tau(2^{p-1}) = \tau(2a_{2^{p-2}})
   \]
   Since \( \tau(2^{p-1}) = p \) (because \( 2^{p-1} \) has \( p \) divisors), we get:
   \[
   \tau(2a_{2^{p-2}}) = p
   \]
   Since \( p \) is prime, the only integers \( n \) for which \( \tau(n) = p \) are of the form \( q^{p-1} \) for some prime \( q \). Therefore:
   \[
   2a_{2^{p-2}} = 2^{p-1} \implies a_{2^{p-2}} = 2^{p-2}
   \]
   which proves the claim. \(\blacksquare\)

2. **Claim 2:** If \( n \) is a positive integer and \( p \) is prime such that \( p > n \ge 2 \), then:
   \[
   a_{p-n} + a_n = p
   \]

   **Proof:** From \( P(p-n, n) \), we have:
   \[
   \tau(p) = \tau(a_{p-n} + a_n)
   \]
   Since \( \tau(p) = 2 \) (because \( p \) is prime), we get:
   \[
   \tau(a_{p-n} + a_n) = 2
   \]
   This implies that \( a_{p-n} + a_n \) is a prime number.

3. **Subclaim 2.1:** If \( n \) is a positive integer and \( p \) and \( q \) are primes such that \( p > q > n \ge 2 \), then:
   \[
   a_{p-n} \ne a_{q-n}
   \]

   **Proof:** Suppose there exists \( n, p, q \) such that \( a_{p-n} = a_{q-n} \). Then this means \( a_{p-n} + a_n = a_{q-n} + a_n = r \) for some prime \( r \).

   Since \( p > q \ge 3 \), this means there is a positive integer \( k \) which lies strictly between \( p \) and \( q \), where \( k \) is composite. Since \( a_1, a_2, a_3, \ldots \) is an increasing sequence, this means that \( a_{k-n} + a_n = r \), which implies that \( a_{k-n} + a_n \) is prime.

   However, from \( P(k-n, n) \):
   \[
   \tau(k) = \tau(a_{k-n} + a_n) \implies \tau(k) = 2
   \]
   which implies that \( k \) is prime, which is a contradiction. \(\blacksquare\)

4. Now, we prove Claim 2. Let \( m \) be the smallest prime such that \( 2^{m-2} > p \), and let \( S \) be the set of all positive integers less than \( 2^{m-2} \) which are of the form \( q-n \) for some prime \( q \), and \( T \) be the set \( \{s_k \mid k \in S\} \).

   Since \( s_k < s_{2^{m-2}} = 2^{m-2} \), this means all the elements of \( T \) must be less than \( 2^{m-2} \) as well. We know that all of the elements from \( T \) must be of the form \( p-n \) as well, which implies that \( T \subseteq S \).

   Furthermore, according to Subclaim 2.1, we find that \( a_{p-n} \ne a_{q-n} \) for any primes \( p \) and \( q \) satisfying \( p > q > n \ge 2 \). This implies that all the elements of \( T \) are distinct. This shows that \( |S| = |T| \), which implies that \( S = T \).

   Therefore, \( a_{p-n} = p-n \) for all \( n \) and prime \( p \) satisfying \( p > n \ge 2 \). \(\blacksquare\)

5. Finally, we use induction to complete the proof that \( a_n = n \) for all positive integers \( n \).

   The base cases are when \( n = 1 \) and \( n = 2 \), which easily follows from Claim 1.

   For the inductive step, suppose that we know that \( a_n = n \) for all \( n = 1, 2, 3, \ldots, k \).

   According to Bertrand's Postulate, there exists a prime \( p \) such that \( k < p < 2k \). Also, \( p - k < 2k - k = k \). From our induction hypothesis, this implies that \( a_{p-k} = p - k \). Hence, from Claim 2:
   \[
   a_k + a_{p-k} = p \implies a_k + (p - k) = p \implies a_k = k
   \]
   which completes the proof. \(\blacksquare\)

The final answer is \( \boxed{ a_n = n } \) for all positive integers \( n \).