I have the following ZIO program with two processes that both run forever:
    for {
      ..
      numberProvider <- numberProvider(queue).fork  // runs forever
      numberService <- numberService(queue)         // runs forever
      ..
    } yield ()
The above code works, but I was wondering if this is good practice.
There are 2 questions:
Is it ok, to run the 2. process on the main program. Or should it be also a Fiber?
Do I have to join the Fibers in the end, even if they run forever and therefore never reach the join?
for {
  ..
  numberProvider <- numberProvider(queue).fork  // runs forever
  numberService <- numberService(queue)         // runs forever
  ..
  _ <- numberProvider.join // join in any case
} yield ()
you don't have to .join fiber if they run forever. 
Note that since 1.0.0-RC17, #zio added the .daemon combinator for exactly that reason, see release note here: https://github.com/zio/zio/releases/tag/v1.0.0-RC17 and that from now on, .fork should be avoided for forever-running fibers. 
From the answer of fanf42 I adjusted my code to:
for {
      ..
      numberProvider <- numberProvider(queue).daemon  // runs forever
      numberService <- numberService(queue)           // runs forever
      ..
    } yield ()
But this did not work (changed fork to daemon). It never got to the next line.
So make sure to fork the daemon as well.
for {
      ..
      numberProvider <- numberProvider(queue).daemon.fork  // runs forever
      numberService <- numberService(queue)                // runs forever
      ..
    } yield ()
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