The problem is:
A class with $2N$ students took a quiz, on which the possible scores were $0,1,\dots,10$. Each of these scores occurred at least once, and the average score was exactly $7.4$ Show that the class can be divided into two groups of N students in such a way that the average score for each group is exactly $7.4$.
The provided solution on the Putnam website, after defining $s_i = \sum_{j=i+1}^{i+N} a_j$, where $a_1, \dots , a_{2N}$ are the scores in increasing order and some calculations, states that
$$a_i < s_i +a_{i+N+1}-7.4N \le a_{i+N+1}$$
then it immediately concludes by saying that we can thus find $N$ scores that sum to $7.4N$ by taking $a_i, \dots, a_{i+1+N}$ and subtracting the value $s_i +a_{i+N+1}-7.4N $,am I getting this right?
I don't understand why this would work given the equality above. Also how can we be sure that $s_i +a_{i+N+1}-7.4N$ is a score value that we can subtract out? there must be something I am not understanding.
The solution is quite specific about the $i$ that you choose in order to make the inequalities hold: it is the largest such that $s_i\leq 7.4N$. let us call $A$ the set $\{a_i,\dots,a_{i+N+1}\}$ So if $s_i=7.4N$, $s_i-7.4N+a_{i+N+1}=a_{i+N+1}$, so obviously you can remove $a_{i+N+1}$ from $A$ and obtained the desired set. If instead $s_i<7.4N$, you know that $s_i+a_{i+N+1}>s_{i+1}>7.4N$ by the choice of $i$, and you are sure that $7.4N$ is an integer because $7.4=37/5$, and since $7.4\cdot 2N$ is an integer you know that $5|N$. So $s_i-7.4N+a_{i+N+1}$ is an integer and it is obviously less than $10$ and larger than $a_i$. But then, by the hypothesis that every value is taken, you can find in $A$ the corresponding element and remove it.