Find all pairs  $(m,n)$ of positive odd integers, such that $n \mid 3m+1$ and $m \mid n^2+3$.
To find all pairs \((m,n)\) of positive odd integers such that \(n \mid 3m+1\) and \(m \mid n^2+3\), we start by analyzing the given conditions.

1. **Initial Observations:**
   - Since \(m\) and \(n\) are odd, both \(3m+1\) and \(n^2+3\) are even.
   - Therefore, we can rewrite the conditions as:
     \[
     2n \mid 3m+1 \quad \text{and} \quad 4m \mid n^2+3
     \]

2. **Assumptions and Inequalities:**
   - Assume \(\frac{3m+1}{2n} \ge 3\) and \(\frac{n^2+3}{4m} \ge 3\).
   - This implies:
     \[
     3m+1 \ge 6n \quad \text{and} \quad n^2+3 \ge 12m
     \]

3. **Combining Conditions:**
   - The problem conditions imply:
     \[
     mn \mid (3m+1)(n^2+3)
     \]
   - Expanding and simplifying:
     \[
     mn \mid n^2 + 9m + 3
     \]
   - Using the inequalities:
     \[
     \frac{6n-1}{3}n \le mn \le n^2 + 9m + 3 \le n^2 + 9\frac{n^2+3}{12} + 3
     \]
   - Simplifying further:
     \[
     3n^2 - 4n - 63 \le 0
     \]
   - Solving the quadratic inequality:
     \[
     n \le 5
     \]

4. **Case Studies:**
   - We now check the possible values of \(n = 1, 3, 5\).

   **Case 1: \(n = 1\)**
   - \(3m + 1 = 2n \implies 3m + 1 = 2 \implies m = \frac{1}{3}\) (not possible since \(m\) must be an integer).

   **Case 2: \(n = 3\)**
   - \(3m + 1 = 2n \implies 3m + 1 = 6 \implies m = \frac{5}{3}\) (not possible since \(m\) must be an integer).

   **Case 3: \(n = 5\)**
   - \(3m + 1 = 2n \implies 3m + 1 = 10 \implies m = 3\)
   - Check if \(m = 3\) satisfies \(m \mid n^2 + 3\):
     \[
     3 \mid 5^2 + 3 \implies 3 \mid 28 \quad (\text{not true})
     \]

5. **Further Case Studies:**
   - **Case 1: \(3m + 1 = 2n\)**
     - \(m \mid n^2 + 3 \implies m \mid \left(\frac{3m+1}{2}\right)^2 + 3 \implies m \mid 13 \implies m = 1, 13\)
     - For \(m = 13\), \(n = 20\) (not possible since \(n\) must be odd).
     - For \(m = 1\), \(n = 2\) (not possible since \(n\) must be odd).

   - **Case 2: \(3m + 1 = 4n\)**
     - \(m \mid n^2 + 3 \implies m \mid \left(\frac{3m+1}{4}\right)^2 + 3 \implies m \mid 49 \implies m = 1, 7, 49\)
     - For \(m = 49\), \(n = 37\) (valid solution).
     - For \(m = 1\), \(n = 1\) (valid solution).

   - **Case 3: \(n^2 + 3 = 4m\)**
     - \(n \mid 3m + 1 \implies n \mid 3\frac{n^2 + 3}{4} + 1 \implies n \mid 13 \implies n = 1, 13\)
     - For \(n = 13\), \(m = 43\) (valid solution).

   - **Case 4: \(n^2 + 3 = 8m\)**
     - Not possible as LHS is \(\equiv 4 \pmod{8}\).

Combining these, we get the solutions:
\[
\boxed{(m,n) = (1,1), (43,13), (49,37)}
\]