N/1 + N/2 + N/3 + ... + N/N

$$$T(n, x)$$$ — expected number of balls to be thrown so that each of $$$n$$$ bins has at least one ball and there are $$$x$$$ empty bins.

Answer for the problem is $$$T(n, n)$$$.

This is known as the coupon collector's problem (https://en.wikipedia.org/wiki/Coupon_collector%27s_problem#Calculating_the_expectation)

consider an array $$$a$$$ where $$$a_x$$$ is the expected number of balls to be thrown where $$$x$$$ bins are empty and $$$n-x$$$ bins has at least one ball.

let's start from $$$x=n$$$ to $$$x=0$$$. we can notice that $$$a_n = 0$$$ because the expected number of balls to be thrown when all the bins has at least one ball is $$$0$$$.

now $$$a_x = 1 + \frac{x}{n} \times a_x + \frac{n-x}{n} \times a_{x+1}$$$, that's because there is a probability equals $$$\frac{x}{n}$$$ that you'll throw a ball into a bin that has at least one ball already and a probability equals $$$\frac{n-x}{n}$$$ that you'll throw a ball into an empty bin. plus one for the ball you are throwing now.

now for the previous equation, we can modify it so that $$$a_x = \frac {1 + \frac{n-x}{n} \times a_{x+1}} { 1 - \frac{x}{n}}$$$. now your answer is $$$a_0$$$. I hope everything I said is clear