4. The maximum two-digit prime factor of the integer $n=\mathrm{C}_{200}^{100}$ is ( ).
(A) 61
(B) 67
(C) 83
(D) 97
4. (A).

Let $p$ be the largest two-digit prime factor of $n=\mathrm{C}_{200}^{100}=\frac{200!}{(100!)^{2}}$, then $0<p<100$.

Since $(100!)^{2}$ contains $p^{2}$, 200! must contain at least $p^{3}$. Therefore, $3 p<200$, which means $p<\frac{200}{3}<67$, yielding $p=61$.