2. Given positive integers $a, b (a>b)$ such that the sum of $a+b$, $a-b$, $a b$, and $\frac{a}{b}$ is 243. Then the number of pairs $(a, b)$ with this property ( ).
(A) has exactly 1 pair
(B) has exactly 2 pairs
(C) has a finite number of pairs, but more than 1 pair
(D) does not exist
2. B.

From the given, we have
$$
(a+b)+(a-b)+a b+\frac{a}{b}=243,
$$

which simplifies to $2 a+a b+\frac{a}{b}=243$.
Since $a, b, 243$ are all positive integers, $a$ must be a multiple of $b$.
Let's assume $a=b k$, substituting into equation (1) we get
$$
2 b k+b^{2} k+k=243 \text {. }
$$

Completing the square, we have $(b+1)^{2}=\frac{243}{k}=\frac{9 \times 27}{k}$.
Therefore, for this equation to hold, we can only have
$$
k=27, b=2, a=54 ; k=3, b=8, a=24 .
$$

Thus, the pairs that satisfy the condition are only $(54,2),(24,8)$.