Let $\mathbb{N}$ be the set of positive integers. Across all function $f:\mathbb{N}\to\mathbb{N}$ such that $$mn+1\text{ divides } f(m)f(n)+1$$ for all positive integers $m$ and $n$, determine all possible values of $f(101).$
1. **Assertion and Initial Setup:**
   Let \( P(m, n) \) be the assertion \( mn + 1 \mid f(m)f(n) + 1 \) for all positive integers \( m \) and \( n \).

2. **Assumption and Contradiction:**
   Assume for the sake of contradiction that there exists a prime number \( p \neq 101 \) such that \( p \mid f(101) \).

3. **Existence of \( a \):**
   Since \( \gcd(101, p) = 1 \), by Bezout's identity, there exists a positive integer \( a \) such that \( a \equiv \frac{-1}{101} \pmod{p} \). This means \( 101a \equiv -1 \pmod{p} \).

4. **Applying Assertion \( P(a, 101) \):**
   From \( P(a, 101) \), we have:
   \[
   101a + 1 \mid f(a)f(101) + 1
   \]
   Since \( 101a \equiv -1 \pmod{p} \), it follows that \( 101a + 1 \equiv 0 \pmod{p} \), so \( p \mid 101a + 1 \).

5. **Divisibility Condition:**
   Therefore, \( p \mid f(a)f(101) + 1 \). Given \( p \mid f(101) \), we can write \( f(101) = pk \) for some integer \( k \). Thus:
   \[
   p \mid f(a)pk + 1
   \]
   Simplifying, we get:
   \[
   p \mid (f(a)k)p + 1 \implies p \mid 1
   \]
   This is a contradiction since \( p \) is a prime number and cannot divide 1.

6. **Conclusion from Contradiction:**
   Thus, \( f(101) \) cannot have any prime divisors other than 101. Therefore, \( f(101) \) must be of the form \( 101^k \) where \( k \in \mathbb{N} \).

7. **Verification with \( P(101, 101) \):**
   Applying \( P(101, 101) \):
   \[
   101^2 + 1 \mid f(101)f(101) + 1 \implies 101^2 + 1 \mid 101^{2k} + 1
   \]
   We need:
   \[
   101^{2k} + 1 \equiv 0 \pmod{101^2 + 1}
   \]
   Simplifying, we get:
   \[
   101^{2k} \equiv -1 \pmod{101^2 + 1}
   \]
   This implies:
   \[
   (101^2)^k \equiv -1 \pmod{101^2 + 1}
   \]
   Since \( 101^2 \equiv -1 \pmod{101^2 + 1} \), we have:
   \[
   (-1)^k \equiv -1 \pmod{101^2 + 1}
   \]
   Therefore, \( k \) must be odd.

8. **General Form of \( f(x) \):**
   Clearly, \( f(x) = x^k \) for all \( x \in \mathbb{N} \) where \( k \) is an odd positive integer satisfies the condition.

The final answer is \( \boxed{ f(101) = 101^k } \) for every odd positive integer \( k \).