Let $S_i$ be the number appearing on the board on the $i$th minute. Then $S_{i+1} = S_i(S_i + 1)$, so $S_{i+1}$ contains all prime divisors of $S_i$ plus at least one more.

Let $S_1, \dots, S_{100}$ be the numbers that were written on the board in the first $100$ minutes. Suppose that before writing the number $S_i$ on the board, the numbers $a_1, \dots, a_k$ were present. Then $S_i = a_1^2 + a_2^2 + \dots + a_k^2$, and the next number written is $S_{i+1} = a_1^2 + a_2^2 + \dots + a_k^2 + S_i^2 = S_i^2$.

Thus, $S_{i+1} = S_i(S_i + 1)$. Therefore, $S_{i+1}$ contains in its prime factorization all the prime numbers that divide $S_i$, plus at least one new prime divisor (a divisor of $1 + S_i$). Since $S_1 > 1$, $S_1$ contains at least one prime divisor in its factorization. Hence, by induction, for $i = 1, 2, \dots, 100$, the number $S_i$ contains at least $i$ distinct prime divisors in its factorization.