jmc

Let the two numbers be $m=AB$ and $n=CD$ (where $A,B,C$ and $D$ are digits). The average of $m$ and $n$ is $\frac{m+n}{2}$ and the number formed by writing $m$ before the decimal point and $n$ after the decimal point is: $$AB.CD=AB+0.CD=AB+\frac{CD}{100}=m+\frac{n}{100}.$$ Setting these equal gives: $$\begin{array}{rl}\frac{m+n}{2}& =m+\frac{n}{100}\\ 50m+50n& =100m+n\\ 49n& =50m\end{array}$$ From this it follows that $n$ a multiple of 50. As $n$ is a 2-digit positive integer this means that $n=50$. So now $50m=49n=49\cdot 50$, so $m=49$. Thus the integers are $49$ and $50$, so the smaller integer is $\boxed{49}$.