The short answer is that, on average, the change in momentum of the clown/balls system is zero, and so the net external force acting on the system is zero.  The only external forces are the weights and the force from the scale, so
[F_SCALE - M_CLOWNg - 5m_BALLg]Dt = Dp = 0.
F_SCALE = M_CLOWNg + 5m_BALLg.

Let T be the time between successive catches of a given ball; this ball is then in the air a time 4T/5 and in the clown's hand for T/5 (as indeed are the others).  The ball's momentum at the beginning of this interval and at the end of this interval are the same, so Dp_BALL = 0.
Look at just one ball.  Consider the impulse that acts on the ball over this cycle: the weight acts over the whole interval and a force from the clown's hand acts over a fifth of the cycle.  We'll assume a constant force, although we could make use of the time-averaged force instead.
-m_CATgT + FT/5 = Dp_BALL= 0.
F = 5m_BALLg.
By the third law, this is also the force the one ball exerts on the clown while it is in his hand (for time interval T/5).  Since the clown is not going anywhere, the forces (or impulses) on him must also sum to zero, so let's use NII:
N_SCALE - m_CLOWNg - F = m_CLOWNa = 0
N_SCALE = m_CLOWNg + F = m_CLOWNg + 5m_BALLg.
This will also be true when the next ball is in his hand.

What if the clown tosses the balls such that he is occasionally empty-handed?  Let t be the time any given ball spends in his hand.  Then that ball spends time T - t in the air.  The ball's impulse equation becomes:
-m_BALLgT + Ft = Dp_BALL = 0.
F = m_BALLgT/t .
The shorter the time interval he handles each ball, the larger the force he must apply to it and by exactly the same fraction.  The average force, though, will then be the same as it was above.

A similar argument works if he holds more than one ball at a time.