How would you interpret this problem? *
Let $h$ and $n$ be the number of hits and number of at-bats after hitting the single, respectively. The answer to the problem suggests solving from the following identity:
$$\frac{h - 1}{n - 1} + \frac{1}{100} = \frac{h}{n}$$
I started from:
$$\lfloor \frac{h-1}{n - 1} \ 1000 \rfloor + 10 = \lfloor {\frac{h}{n} \ 1000} \rfloor$$
Which seems much harder to solve. I don't know how to do it in general. Computer search counted 336 pairs $(h, n)$ with $h$ between 2 and 30.
(*) "Which Way Did the Bicycle Go?", Kornhauser et al., 1996, p. 34
We have $$\frac{h - 1}{n - 1} + \frac{1}{100} = \frac{h}{n}\\100n(h-1)+n(n-1)=100h(n-1)\\ h=n-\frac {n(n-1)}{100}$$ So we need $n(n-1)$ to be a multiple of $100$. One of $n,n-1$ needs to be a multiple of $25$ and the other a multiple of $4$. This gives $n=25, h=19$
He started at $\frac {18}{24}=0.750$ and ended at $\frac {19}{25}=0.760$. Quite a hitter. The next solution has $n=75$ and it turns out that also works with $18$ hits, with the average going from $0.240$ to $0.250$. In either case he started with $18$ hits. Thanks to paw88789 for catching my error