A prime number $p$ is mundane if there exist positive integers $a$ and $b$ less than $\tfrac{p}{2}$ such that $\tfrac{ab-1}{p}$ is a positive integer. Find, with proof, all prime numbers that are not mundane.
To find all prime numbers that are not mundane, we need to determine the primes \( p \) for which there do not exist positive integers \( a \) and \( b \) less than \( \frac{p}{2} \) such that \( \frac{ab-1}{p} \) is a positive integer. This is equivalent to finding primes \( p \) for which there do not exist \( a \) and \( b \) such that \( ab \equiv 1 \pmod{p} \).

1. **Check small primes:**
   - For \( p = 2 \):
     \[
     \left(1, \frac{2}{2}\right) = (1, 1)
     \]
     There are no integers in this range, so \( p = 2 \) is not mundane.
   
   - For \( p = 3 \):
     \[
     \left(1, \frac{3}{2}\right) = (1, 1.5)
     \]
     There are no integers in this range, so \( p = 3 \) is not mundane.
   
   - For \( p = 5 \):
     \[
     a, b \in \left(1, \frac{5}{2}\right) = (1, 2.5)
     \]
     The only possible values are \( a = b = 2 \). However,
     \[
     2 \cdot 2 = 4 \not\equiv 1 \pmod{5}
     \]
     So, \( p = 5 \) is not mundane.
   
   - For \( p = 7 \):
     \[
     a, b \in \left(1, \frac{7}{2}\right) = (1, 3.5)
     \]
     The possible values are \( a, b \in \{2, 3\} \). Checking all combinations:
     \[
     2 \cdot 2 = 4 \not\equiv 1 \pmod{7}
     \]
     \[
     2 \cdot 3 = 6 \not\equiv 1 \pmod{7}
     \]
     \[
     3 \cdot 3 = 9 \equiv 2 \not\equiv 1 \pmod{7}
     \]
     So, \( p = 7 \) is not mundane.
   
   - For \( p = 13 \):
     \[
     a, b \in \left(1, \frac{13}{2}\right) = (1, 6.5)
     \]
     The possible values are \( a, b \in \{2, 3, 4, 5, 6\} \). Suppose \( 13 \) is mundane, then there exist \( a, b \) such that \( ab = 13k + 1 \) for some \( k \). Checking for \( k = 1 \):
     \[
     ab = 14
     \]
     None of \( \{2, 3, 4, 5, 6\} \) multiply to 14. Checking for \( k = 2 \):
     \[
     ab = 27
     \]
     None of \( \{2, 3, 4, 5, 6\} \) multiply to 27. For \( k \geq 3 \):
     \[
     36 \geq ab = 13k + 1 \geq 40
     \]
     This is a contradiction. So, \( p = 13 \) is not mundane.

2. **Check larger primes:**
   - For \( p = 11 \):
     \[
     a, b \in \left(1, \frac{11}{2}\right) = (1, 5.5)
     \]
     The possible values are \( a, b \in \{2, 3, 4, 5\} \). Checking:
     \[
     3 \cdot 4 = 12 \equiv 1 \pmod{11}
     \]
     So, \( p = 11 \) is mundane.

3. **General case for \( p \geq 19 \):**
   - Assume \( p \geq 19 \) is not mundane. We use induction to show \( p \equiv 1 \pmod{2^k} \) for all \( k \) such that \( 2^k < \frac{p}{2} \).
   - Base case \( k = 1 \):
     \[
     p \text{ is odd, so } p \equiv 1 \pmod{2}
     \]
   - Inductive step: Assume \( p \equiv 1 \pmod{2^{r-1}} \). We show \( p \equiv 1 \pmod{2^r} \). If not, then \( p \equiv 2^{r-1} + 1 \pmod{2^r} \). Then,
     \[
     (2^{r-1} - 1)p + 1 \equiv (2^{r-1} - 1)(2^{r-1} + 1) + 1 \equiv 2^{2(r-1)} \equiv 0 \pmod{2^r}
     \]
     This implies \( \frac{(2^{r-1} - 1)p + 1}{2^r} \) is an integer. Since \( 1 < 2^r < \frac{p}{2} \), this contradicts \( p \) not being mundane. Thus, \( p \equiv 1 \pmod{2^r} \).

   - Let \( r \) be such that \( 2^r < \frac{p}{2} < 2^{r+1} \). Then \( p \equiv 1 \pmod{2^r} \) and \( p < 2^{r+2} \). So, \( p = 2^r + 1 \), \( 2^{r+1} + 1 \), or \( 3 \cdot 2^r + 1 \). If \( p \equiv 2 \pmod{3} \), then \( \frac{p+1}{3} \) is an integer and \( 1 < \frac{p+1}{3} < \frac{p}{2} \), contradicting \( p \) not being mundane. Thus, \( p \equiv 1 \pmod{3} \), and \( p \neq 2^r + 1 \) or \( 2^{r+1} + 1 \).

The final answer is \( p = \boxed{2, 3, 5, 7, 13} \)