In a list of $N$ non-negative integers whose sum is $100,$ the integer $3$ occurs more often than any other. What is the largest possible value of $N?$

Why not find a valid list and try to increase the number of integers in the list?

What are the sensible integers in the largest list?

… remember that you can use $0$ as well.

… for example, having $5$ is not sensible because we can have $3,$ $2$ and $0$ instead and have more integers.

How about expressing the number of other sensible integers in terms of the number of $3$s to find $N?$

Have you considered all of the constraints given?

Having numbers greater than $3$ in the list is not useful, since we can have more integers for the same value by splitting that number into some $3$s, $1$s and $0$s. A $2$ can be converted into two $1$s and two $2$s can be converted into $3,1$ and $0.$

Since any two $2$s can be converted to $3,1$ and $0,$ which doesn’t affect our quota of $1$s and $0$s (bounded by number of $3$s), the optimal solution will have at most one $2.$ Although there is an optimal solution that uses $2$ (see below), it is possible to reach optimum without $2$s. So for now, the only numbers we will consider are $0, 1,$ and $3.$

We also notice that we can exchange a $3$ and a $0$ for 3 $1$s. Hence our strategy is to maximize the number of $3$s such that we can fit as many $1$s without making the $1$s more common. We then add $0$s up to the number of $3$s. Let $t$ be the number of $3$s which means we’ll have $100-3t$ of $1$s and $t-1$ of $0$s. In total, we’ll have $N=t+100-3t+t-1=100-t-1,$ conditioned by $t>100-3t,$ which gives $t=26$ and $N=73.$

The two solutions are 26 $3$s, 22 $1$s, 25 $0$s or 25 $3$s, 1 $2$, 23 $1$s, 24 $0$s.