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Solution

Given: $\sum x_i^2 = 400$ and $\sum x_i = 80$
Using the inequality:
$\sum x_i^2 \geq \dfrac{(\sum x_i)^2}{n}$
This follows from Cauchy-Schwarz inequality:
$n \cdot \sum x_i^2 \geq \left(\sum x_i\right)^2$
$\Rightarrow n \times 400 \geq (80)^2$
$\Rightarrow 400n \geq 6400$
$\Rightarrow n \geq \dfrac{6400}{400}$
$\Rightarrow n \geq 16$
Also, Variance $\geq 0$:
$\dfrac{\sum x_i^2}{n} - \left(\dfrac{\sum x_i}{n}\right)^2 \geq 0$
$\Rightarrow \dfrac{400}{n} - \dfrac{6400}{n^2} \geq 0$
$\Rightarrow \dfrac{400n - 6400}{n^2} \geq 0$
$\Rightarrow n \geq 16$
$\therefore$ A possible value of $n$ is $\boxed{n = 20}$