, 2015

We are given that three consecutive odd integers have an average of 7.

These three integers must be 5, 7 and 9.

One way to see this is to let the three integers be $a-2,a,a+2$. (Consecutive odd integers differ by 2.)

Since the average of these three integers is 7, then their sum is $3\cdot 7 = 21$.

Thus, $(a-2)+a+(a+2)=21$ or $3a=21$ and so $a=7$.

When $m$ is included, the average of the four integers equals their sum divided by 4, or $\dfrac{21+m}{4}$.

This average is an integer whenever $21+m$ is divisible by 4.

Since 21 is 1 more than a multiple of 4, then $m$ must be 1 less than a multiple of 4 for the sum $21+m$ to be a multiple of 4.

The smallest positive integers $m$ that are 1 less than a multiple of 4 are 3, 7, 11, 15, 19.

Since $m$ cannot be equal to any of the original three integers 5, 7 and 9, then the three smallest possible values of $m$ are 3, 11 and 15.

The sum of these possible values is $3+11+15=29$.