Answer: A

Solution:

Subtracting Zelda's first-semester average from all four semester averages does not affect the comparison. We may therefore write the semester averages as $0,18$ for Zelda and $3,21$ for Yolanda. Let $p$ and $q$ be the fractions of Yolanda's and Zelda's quizzes, respectively, that occurred in the second semester. Their yearly-average difference is $$(3+18p)-18q=3+18(p-q).$$ Because $0\lt p,q\lt 1,$ this difference is less than $21,$ so it cannot be $22.$

The other options really can occur. Taking $(p,q)=(\tfrac14,\tfrac34)$ makes Zelda's average higher; taking $p=q$ makes Yolanda's average $3$ points higher (and also verifies the last option after adding $3$ to every Zelda score); and taking $(p,q)=(\tfrac13,\tfrac12)$ makes the yearly averages equal. Each displayed fraction can be realized by positive integer quiz counts.

Thus, the correct answer is A.