Here is a remark from Silverman's Arithmetic of Elliptic Curves:
[in his notation, $K$ is a number field and $M_K^0$ is the set of non-archimedian places in $K$]
How does he know that $v(a_1),...,v(a_6)\geq 0$ and $v(\Delta)=0$ for all but finitely many $v\in M_K^{0}$? And why is this a minimal Weierstrass equation?
This is basically because an integer can only have finitely many prime factors. I'll take $K=\Bbb Q$. Otherwise one has to replace prime factors by prime ideal factors, but the principle is the same.
One can write $a_i=b_i/c_i$ where $b_i$ and $c_i$ are integers, and $c_i\ne0$. Then $v(c_i)<0$ if $v=v_p$ with $p$ a prime factor of $c_i$. There are only finitely many such $p$. Also $\Delta=u/v$ with $u$ and $v$ nonzero integers, and $v(\Delta)\ne0$ only if $v=v_p$ for $p\mid uv$. Again there are only finitely many such $p$.
When the conditions holds, $v(\Delta)=0$ and more-or-less by definition the equation is minimal Weierstrass over the field $K_v$. It maybe not a minimal Weierstrass equation over the number field $K$.