jmc

Let the first odd integer be $a$. Let the rest of the odd integers be $a+2,a+4,a+6,\dots ,a+2(n-1)$, for a total of $n$ integers. The arithmetic mean of these integers is equal to their sum divided by the number of integers, so we have $$y=\frac{na+(2+4+6+\cdots +2(n-1))}{n}$$ Notice that $2+4+6+\cdots +2(n-1)=2(1+2+3+\cdots +n-1)=2\frac{(n-1)(n-1+1)}{2}=n(n-1)$. Substituting and multiplying both sides by $n$ yields $$yn=na+n(n-1)$$ Dividing both sides by $n$, we have $$y=a+n-1$$ The sum of the smallest and largest integers is $a+a+2(n-1)$, or $2a+2(n-1)=2(a+n-1)=2y$.

Hence the answer is $\boxed{2y}$.