1. We are given $\sum_{i=1}^n x_i^2 = 400, \qquad \sum_{i=1}^n x_i = 80.$ We need to find which among the given values of $n$ is possible.

2. Use the standard inequality: $\left(\sum_{i=1}^n x_i\right)^2 \le n \sum_{i=1}^n x_i^2.$ This is from Cauchy–Schwarz.

3. Substitute the given values: $80^2 \le n \cdot 400$ $6400 \le 400n$ $n \ge 16.$

So, any possible value of $n$ must satisfy $n \ge 16.$

4. Now check the options:

• $18$: satisfies $18 \ge 16$ ✔️
• $15$: does not satisfy ✖️
• $12$: does not satisfy ✖️
• $9$: does not satisfy ✖️

So only $n=18$ can be possible.

5. We can also verify existence for $n=18$. We need observations with sum $80$ and sum of squares $400$. The average is $\bar x = \frac{80}{18} = \frac{40}{9}.$ Since equality in Cauchy would require all values equal, and here $18>16$, such a set can exist with some variation. Hence $n=18$ is indeed possible.

Therefore, the correct option is $\boxed{18}.$