Solution:

Assume that $b$ is the larger of the two numbers. Then $b^{2}$ and $a^{2}$ end in the same last two digits, so $b^{2}-a^{2}$ is a multiple of $100$. That is, using the difference of squares to factor, $(b+a) \cdot (b-a)$ is a multiple of $100$, and hence also a multiple of both $4$ and $25$.

For the product to be a multiple of $25$, either at least one of the numbers is a multiple of $25$, or both are multiples of $5$. If they are both multiples of $5$, then $a$ must be a multiple of $5$, so $a^{2}$ ends in $25$ or $00$. There are many possibilities of this kind: Ending in $25$, we have $5$ and $15$, $5$ and $25$, $5$ and $35$, and so on, up through $35$ and $45$, giving averages of $10, 15, 20, 25, 30, 35$, and $40$. Ending in $00$, we have $10$ and $20$, and so on, up through $30$ and $40$, which gives the same list of averages. (We did give credit if you included $50$ and thus put $45$ in your list of possible averages.)

Now let's consider the possibility that one of the numbers is a multiple of $25$. Neither $a-b$ nor $a+b$ could equal $25$ or $75$, because if they did, then one of $a$ and $b$ is odd, while the other is even, so their squares can't end in the same digit. Also, $b-a$ can't be $50$, because $b<50$. So the only way to obtain a multiple of $25$ is for $b+a=50$, in which case the average is $25$.

Of course, you could also simply list all the squares from $1^{2}$ through $49^{2}$ and find all the pairs with the same last two digits!