Find all positive integers $n$ for which the largest prime divisor of $n^2+3$ is equal to the least prime divisor of $n^4+6.$
1. Let \( p \) be the largest prime divisor of \( n^2 + 3 \) and also the smallest prime divisor of \( n^4 + 6 \). Since \( n^4 + 6 = (n^2 + 3)(n^2 - 3) + 15 \), we get that \( p \mid 15 \).

2. The prime divisors of 15 are 3 and 5. We will consider each case separately.

3. **Case 1: \( p = 3 \)**
   - Since \( p \) is the smallest prime divisor of \( n^4 + 6 \), then 2 does not divide \( n^4 + 6 \). Hence \( n \equiv 1 \pmod{2} \), meaning \( n \) is odd.
   - Furthermore, since \( p \) is the largest prime divisor of \( n^2 + 3 \), we have \( n^2 + 3 = 2^a 3^b \) for some natural numbers \( a \) and \( b \).
   - Consider \( n^2 + 3 \pmod{8} \). Since \( n \equiv 1 \pmod{2} \), \( n^2 \equiv 1 \pmod{8} \). Thus, \( n^2 + 3 \equiv 4 \pmod{8} \), which means \( a = 2 \) (since \( 2^2 = 4 \)).
   - Since \( p \mid n^2 + 3 \), then \( 3 \mid n \). Let \( n = 3k \) for some integer \( k \).
   - Substitute \( n = 3k \) into \( n^2 + 3 \):
     \[
     (3k)^2 + 3 = 9k^2 + 3
     \]
     - We need \( 9k^2 + 3 = 2^2 \cdot 3 \), so \( 9k^2 + 3 = 12 \). Solving for \( k \):
     \[
     9k^2 + 3 = 12 \implies 9k^2 = 9 \implies k^2 = 1 \implies k = 1
     \]
     - Thus, \( n = 3 \cdot 1 = 3 \).

4. **Case 2: \( p = 5 \)**
   - Note that \( n^2 + 3 \) is never divisible by 5. To see why, consider \( n^2 + 3 \pmod{5} \):
     - If \( n \equiv 0 \pmod{5} \), then \( n^2 + 3 \equiv 3 \pmod{5} \).
     - If \( n \equiv 1 \pmod{5} \), then \( n^2 + 3 \equiv 4 \pmod{5} \).
     - If \( n \equiv 2 \pmod{5} \), then \( n^2 + 3 \equiv 7 \equiv 2 \pmod{5} \).
     - If \( n \equiv 3 \pmod{5} \), then \( n^2 + 3 \equiv 12 \equiv 2 \pmod{5} \).
     - If \( n \equiv 4 \pmod{5} \), then \( n^2 + 3 \equiv 19 \equiv 4 \pmod{5} \).
   - In all cases, \( n^2 + 3 \) is never 0 modulo 5, so \( p \neq 5 \).

Hence, the only solution is \( n = 3 \).

The final answer is \( \boxed{3} \).