Let $S$ be a smooth projective surface of Picard rank $\rho(S)$ over a field $K$, and $\overline{S}$ its algebraic closure.
Take a point $p\in\overline{S}$ and denote by $\overline{X}$ be blow-up of $\overline{S}$ at $p$. Is it possible to chose the point $p$ in such a way that $\overline{X}$ is the algebraic closure of a surface $X$ defined over $K$ and having Picard number equal to $\rho(S)$?
The answer is yes for some $S$ and no for other $S$. I think the answer is no for most $S$ for a reasonable definition of "most".
For a positive example, take $S$ to be $\mathbb P^2$ blown up at a point, of Picard rank $2$. Then $\overline{X}$ will be $\mathbb P^2$ blown up at two points, which is the base change to the algebraic closure of $\mathbb P^2$ blown up at a degree two point, which has Picard rank $2$. (Here I am making the mild assumption that $K$ has a separable quadratic extension.)
For a negative example, take $S$ to be a simple abelian surface with no extra endomorphisms, so its Picard rank is $1$. Then $X$ will always have a nontrivial map defined over $K$ to its Albanese variety (or its minimal model), which means its Picard rank must be at least $2$ (an ample divisor on the target of the map plus a curve must be contracted).
A similar argument should work for any surface of nonnegative Kodaira dimension and Picard rank $1$.
