Parse $2 \times 3 \times 5 \times 7 \times 11=2310$. When $k=12,13, \cdots, 2310+11$, it is not difficult to verify that among $k, k+1, k+2, \cdots, k+99$ (1) at least 76 numbers are divisible by 2, 3, 5, 7, or 11. Therefore, when $k>11$, (1) contains at least 76 composite numbers and at most 24 prime numbers. When $k=2$, (1) contains 26 prime numbers. When $k=1,3,4,5,6,7,8,9,10,11$, (1) contains 25 prime numbers. Therefore, when $k=2$, (1) contains the most prime numbers.
14.4.13 ** Find all integers \( k \geqslant 1 \), such that the sequence \( k, k+1, k+2, \cdots, k+99 \) contains the maximum number of prime numbers.