For the purposes of hypothesis testing, I often find that the simpler approach is the best. 

In this case, I would do exactly what you considered yourself: average all the pre-treatment values and all the post-treatment values for each participant, obtaining two values per participant. **Then you can run a paired t-test on the resulting averages.**

There is *nothing* wrong with this approach. If you do that and you get $p$-value sufficiently low for your purposes, you can call it a day. I guess there are much more complicated mixed models that one can set up here, but I would be skeptical that they produce much lower $p$-values (and if not, then there is no gain). Whereas two **big** advantages of the simple t-test on averages are: (1) it takes five minutes to perform; (2) it takes two lines to explain in a paper.

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PS. If I am not mistaken, then a simple repeated measures ANOVA (that you asked about) cannot be applied in your case.

PPS. Note that without a control group you will not be able to say if the difference between post and pre (in case you observe any) is due to treatment or due to some time passing.

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**Update.** What I wrote above was under assumption that either you have no information about times of individual measurements, or you are happy to assume that the time is irrelevant. @psarka argued (+1) that time of the day is very relevant and, worse, it is unlikely that measurement pre- and post-treatment were distributed along the day in the same way. So if you have the information about measurement times, then you should better take it into account, and the exercise becomes more complicated then. If not, then well, not.

In addition, @robin argued that the day number is important as well, see discussion in the comments.