4. For $\triangle A B C$, the tangent values of the three interior angles are all integers. If similar triangles are counted as only one case, then the total number of cases that meet the conditions is ( ).
(A) impossible
(B) only one
(C) infinitely many
(D) at least two, with a finite number
4.B.

Let $\tan A=x, \tan B=y, \tan C=z, x, y, z$ be non-zero integers, and at most one of them is negative. From the identity
$$
\cot A \cdot \cot B + \cot B \cdot \cot C + \cot C \cdot \cot A = 1,
$$

we get $\frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x} = 1$,
which simplifies to $x y z = x + y + z$.
If $x, y, z$ include a negative integer, without loss of generality, let $z < 0$.
Then, $x y < 1$, which implies $x y \leqslant 0$, a contradiction.
Therefore, $x, y, z$ are all positive integers, and one of them must be 1. Otherwise, if $x, y, z$ are all at least 2, then from equation (1) we have
$$
1 = \frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x} \leqslant \frac{3}{2 \times 2} = \frac{3}{4},
$$

a contradiction.
Assume $x \geqslant y \geqslant z$, then $z = 1$.
From equation (2) we have $x y = x + y + 1$, which simplifies to $(x-1)(y-1) = 2$, giving $x = 3, y = 2$,

Thus, the only solution is $\{\tan A, \tan B, \tan C\} = \{1, 2, 3\}$.