9. Let the first term and common difference of an arithmetic sequence be non-negative integers, the number of terms be no less than 3, and the sum of all terms be $97^{2}$. How many such sequences are there?
9. Let the first term of an arithmetic sequence be $a$, and the common difference be $d$. According to the problem, we have $n a + \frac{n(n-1)}{2} d = 97^2$, which can be rewritten as $[2a + (n-1)d] \cdot n = 2 \times 97^2$. Given that $n$ is a positive integer no less than 3, and 97 is a prime number, the possible values of $n$ can only be $97, 2 \times 97, 97^2$, or $2 \times 97^2$. Since $a > 0$, from $[2a + (n-1)d] \cdot n = 2 \times 97^2$, we know that $2 \times 97^2 \geq n(n-1)d \geq n(n-1)$. Therefore, the only possible value for $n$ is 97, leading to $a + 48d = 97$. For $n = 97$, we have $d = 1, a = 49$ or $n = 97, d = 2, a = 1$.

If $d = 0$, then we have $na = 97^2$, which gives $n = 97, d = 0, a = 97$ or $n = 97^2, d = 0, a = 1$. Therefore, there are 4 such sequences.