Let the rational number $p/q$ be closest to but not equal to $22/7$ among all rational numbers with denominator $< 100$. What is the value of $p - 3q$?
To solve the problem, we need to find the rational number \( \frac{p}{q} \) closest to but not equal to \( \frac{22}{7} \) among all rational numbers with denominator less than 100. We will use the mediant property of fractions, which states that for two fractions \( \frac{a}{b} \) and \( \frac{c}{d} \), the mediant is given by \( \frac{a+c}{b+d} \).

1. **Express \( \frac{22}{7} \) as a mixed number:**
   \[
   \frac{22}{7} = 3 + \frac{1}{7}
   \]
   We need to find the rational number \( \frac{p}{q} \) closest to \( \frac{1}{7} \) with a denominator less than 100.

2. **Find the mediants between \( \frac{0}{1} \) and \( \frac{1}{1} \):**
   \[
   \frac{0}{1} \oplus \frac{1}{1} = \frac{0+1}{1+1} = \frac{1}{2}
   \]
   \[
   \frac{0}{1} \oplus \frac{1}{2} = \frac{0+1}{1+2} = \frac{1}{3}
   \]
   \[
   \frac{0}{1} \oplus \frac{1}{3} = \frac{0+1}{1+3} = \frac{1}{4}
   \]
   Continue this process until we reach \( \frac{1}{7} \):
   \[
   \frac{0}{1} \oplus \frac{1}{4} = \frac{1}{5}
   \]
   \[
   \frac{0}{1} \oplus \frac{1}{5} = \frac{1}{6}
   \]
   \[
   \frac{0}{1} \oplus \frac{1}{6} = \frac{1}{7}
   \]

3. **Case I: \( \frac{p}{q} < \frac{1}{7} \):**
   \[
   \frac{1}{7} \oplus \frac{1}{8} = \frac{1+1}{7+8} = \frac{2}{15}
   \]
   \[
   \frac{2}{15} \oplus \frac{1}{7} = \frac{2+1}{15+7} = \frac{3}{22}
   \]
   Continue this process until the denominator is less than 100:
   \[
   \frac{3}{22} \oplus \frac{1}{7} = \frac{4}{29}
   \]
   \[
   \frac{4}{29} \oplus \frac{1}{7} = \frac{5}{36}
   \]
   \[
   \frac{5}{36} \oplus \frac{1}{7} = \frac{6}{43}
   \]
   \[
   \frac{6}{43} \oplus \frac{1}{7} = \frac{7}{50}
   \]
   \[
   \frac{7}{50} \oplus \frac{1}{7} = \frac{8}{57}
   \]
   \[
   \frac{8}{57} \oplus \frac{1}{7} = \frac{9}{64}
   \]
   \[
   \frac{9}{64} \oplus \frac{1}{7} = \frac{10}{71}
   \]
   \[
   \frac{10}{71} \oplus \frac{1}{7} = \frac{11}{78}
   \]
   \[
   \frac{11}{78} \oplus \frac{1}{7} = \frac{12}{85}
   \]
   \[
   \frac{12}{85} \oplus \frac{1}{7} = \frac{13}{92}
   \]
   \[
   \frac{13}{92} \oplus \frac{1}{7} = \frac{14}{99}
   \]

4. **Case II: \( \frac{p}{q} > \frac{1}{7} \):**
   \[
   \frac{1}{7} \oplus \frac{1}{6} = \frac{2}{13}
   \]
   \[
   \frac{2}{13} \oplus \frac{1}{7} = \frac{3}{20}
   \]
   Continue this process until the denominator is less than 100:
   \[
   \frac{3}{20} \oplus \frac{1}{7} = \frac{4}{27}
   \]
   \[
   \frac{4}{27} \oplus \frac{1}{7} = \frac{5}{34}
   \]
   \[
   \frac{5}{34} \oplus \frac{1}{7} = \frac{6}{41}
   \]
   \[
   \frac{6}{41} \oplus \frac{1}{7} = \frac{7}{48}
   \]
   \[
   \frac{7}{48} \oplus \frac{1}{7} = \frac{8}{55}
   \]
   \[
   \frac{8}{55} \oplus \frac{1}{7} = \frac{9}{62}
   \]
   \[
   \frac{9}{62} \oplus \frac{1}{7} = \frac{10}{69}
   \]
   \[
   \frac{10}{69} \oplus \frac{1}{7} = \frac{11}{76}
   \]
   \[
   \frac{11}{76} \oplus \frac{1}{7} = \frac{12}{83}
   \]
   \[
   \frac{12}{83} \oplus \frac{1}{7} = \frac{13}{90}
   \]
   \[
   \frac{13}{90} \oplus \frac{1}{7} = \frac{14}{97}
   \]

5. **Compare the distances:**
   \[
   \left| \frac{14}{99} - \frac{1}{7} \right| = \left| \frac{14}{99} - \frac{14}{98} \right| = \left| \frac{14}{99} - \frac{14}{98} \right| = \left| \frac{14 \cdot 98 - 14 \cdot 99}{99 \cdot 98} \right| = \left| \frac{14 \cdot (98 - 99)}{99 \cdot 98} \right| = \left| \frac{-14}{99 \cdot 98} \right| = \frac{14}{99 \cdot 98}
   \]
   \[
   \left| \frac{14}{97} - \frac{1}{7} \right| = \left| \frac{14}{97} - \frac{14}{98} \right| = \left| \frac{14 \cdot 98 - 14 \cdot 97}{97 \cdot 98} \right| = \left| \frac{14 \cdot (98 - 97)}{97 \cdot 98} \right| = \left| \frac{14}{97 \cdot 98} \right| = \frac{14}{97 \cdot 98}
   \]

   Since \( \frac{14}{99} \) is closer to \( \frac{1}{7} \) than \( \frac{14}{97} \), we choose \( \frac{14}{99} \).

6. **Calculate \( \frac{p}{q} \) and \( p - 3q \):**
   \[
   \frac{p}{q} = 3 + \frac{14}{99} = \frac{3 \cdot 99 + 14}{99} = \frac{297 + 14}{99} = \frac{311}{99}
   \]
   \[
   p - 3q = 311 - 3 \cdot 99 = 311 - 297 = 14
   \]

The final answer is \( \boxed{14} \).