$A$ PID, $M$ flat (i.e., torsion-free). Then $\operatorname{Ext}_A^1(M,N)$ is injective, for all $N$. Let $A$ be a PID and $M$ a flat (i.e., torsion-free) $A$-module. Then, for every $A$-module $N$, $\text{Ext}_A^1(M, N)$ is injective in $A\text{-}\mathbf{Mod}$.
It is easy when $M$ is finitely generated, since it is free (in particular, projective); thus $\text{Ext}_A^\bullet$-acyclic. Does the general case follow by a filtered colimit argument?
 A: 
Does the general case follow by a filtered colimit argument?
$\newcommand{\Z}{\mathbb{Z}}$
$\newcommand{\Q}{\mathbb{Q}}$
$\newcommand{\R}{\mathbb{R}}$
$\newcommand{\Hom}{\operatorname{Hom}}$
$\newcommand{Ext}{\operatorname{Ext}}$

No.  I spoke about precisely this in class yesterday.  In light of the corresponding properties for the Tor functors and since $\Ext^{\bullet}( \bigoplus_{i \in I} M_i,N) \cong \prod_{i \in I} \Ext^{\bullet}(M_i,N)$, it is natural to wonder about

$\Ext^{\bullet} (\varinjlim M_i,N) \stackrel{?}{\cong} \varprojlim \Ext^{\bullet}(M_i,N)$.

Since for all modules $N$, projective modules are acyclic for $\Hom(\cdot,N)$, if the above "continuity" property of $\Ext$ held true, then all (I mean filtered here) direct limits of projective modules would also be acyclic for $\Hom(\cdot,N)$.  By a famous result of
Govorov-Lazard, these are precisely the flat modules, so we'd have $\Ext^n_R(M,N) = 0$
for all flat $M$, all $N$, and all $n > 0$.
However, there are plenty of counterexamples to this.  My favorite at the moment (as a relative newcomer to this material) is

$\Ext^1_{\Z}(\Q,\Z) \cong \R$.

(Writing $\R$ is a bit "lyrical": we mean of course the additive group of $\mathbb{R}$, which is characterized by being a torsionfree divisible abelian group of cardinality $2^{\aleph_0}$.)
This result is the subject of this short article by James Wiegold.  A second, shorter proof is sketched by J. Rotman in his MathSciNet reiew of Wiegold's article (he says the result is "well known", which indeed seems plausible).  A third (still shorter?) proof is sketched at the end of $\S$ 6.4 of these rough course notes of mine.

Note also that the same section in my notes contains a proof of a mild generalization of your main question: if $R$ is a Dedekind domain, for every torsionfree module $M$ and every module $N$, $\Ext^1_R(M,N)$ is divisible.  (Note that over a Dedekind domain it is true that torsionfree = flat and divisible = injective.  Nevertheless I stand by these two replacements, since the argument actually uses "only" the torsionfree condition and yields "only" the conclusion about divisibility.)
A: You have to prove that $\operatorname{Ext}_A^1(M,N)$ is divisible. Take $a\in A$, $a\neq 0$. Since $M$ is torsion-free we have a short exact sequence $0\to M\stackrel{a\cdot}\to M\to M/aM\to 0$ and this gives rise to a long exact sequence of homology: $\operatorname{Ext}_A^1(M,N)\stackrel{a\cdot}\to\operatorname{Ext}_A^1(M,N)\to\operatorname{Ext}_A^2(M/aM,N)=0$ (the last $\operatorname{Ext}$ is $0$ since a PID has global dimension $\le 1$), and we are done. 
