One-to-one correspondence between these two problems? 
How many 3-digit positive integers are there whose middle digit is
  equal to the sum of the first and last digits?

I noticed that the solution to this problem, $45$, is the same as the solution to the problem

How many 3-digit positive integers are there whose middle digit is the average
  of the first and last digits?

Is this purely a coincidence or is there some sort of a bijection/one-to-one correspondence that links these two?

Note that in both problems, the middle digit (if valid) is determined by the first and last digits.
There are 90 possible pairs of first and last digits $(f,\ell)$, since $1\le f\le 9$ and $0\le\ell\le9$. These can be grouped into 45 pairs $\{ (f,\ell), (10-f,9-\ell) \}$.
Note that exactly one member of each such pair leads to an answer to the first problem: exactly one of $f+\ell$ and $(10-f)+(9-\ell)$ is between 0 and 9, since the two sums add to 19.
Note also that exactly one member of each such pair leads to an answer to the second problem: exactly one of $\frac12(f+\ell)$ and $\frac12((10-f)+(9-\ell))$ is an integer, because the two expressions add to $\frac{19}2$ and both are half-integers.
Therefore the trivial bijection between the sets of 45 pairs induces a bijection between the solution sets of the two problems. For example, $110, 121, 132, 143$ are bijectively mapped to $999, 111, 987, 123$ respectively, while the preimages of $333, 345, 357, 369$ are $363, 385, 792, 770$ respectively.
