Determine minimum number of rows to be reserved. In an arena, each row has $199$ seats. One day, $1990$ students are coming to attend a soccer match. It is only known that at most $39$ students are from the same school. If students from the same school must sit in the same row, determine the minimum number of rows that must be reserved for these students.
My attempt was as follows:
First, because at most $39$ students can come from a school, the minimum number of schools that can come is $ \lceil{ 1990/39} \rceil $, which is equal to $52$. Now, a row can accommodate $ \lfloor {199/39} \rfloor = 5$ schools, so number of required rows should be $ \lceil {52/5} \rceil = 11$ schools.
However, the given solution says that atleast $12$ rows are required. A solution was provided too, but sadly, I couldn't understand a word of it.
Please help me out.
Thank you and stay safe.
 A: In Parcly's now deleted solution, he provides an example where we need 12 rows:   

Consider a situation where there are $56$ schools with $35$ students each and one school with $30$. It is still not possible to have $6$ schools on the same row, but there are $57$ groups in all, so they require at least $12$ rows.

How do we show that 12 rows are sufficient for the arbitrary case?
Hint: You might recognize the "stones of total mass 9 tons, each of at most 1 ton, transported in trucks that can take 3 tons of weight" olympiad problem. This is of a similar flavor, so it's no surprise that we use the Greedy Algorithm, but it involves more work to push through.       

Arrange the schools in decreasing order by the number of students.
Let $s_i$ be the number of students in school $i$, then $s_1 \geq s_2 \geq s_3 \geq \ldots \geq s_I$.
Now, use the greedy algorithm to pack the teams into the first 11 rows, till we cannot fit the very next team.
Say we can only fit up to team $J$, so our hope is that $ s_{J+1} + \ldots + s_I \leq 199$.      
Lemma: Show that $ S_{J+1} \leq 37$.   

 Proof by contradiction. Suppose not, then if the schools that are seated have at least 37 students, then each row has at least $5 \times 37= 185$ students.
 Since $11\times 185 = 2035 > 1990$ hence we have a contradiction.   

Lemma: Show that $s_{J+1} + \ldots + s_I \leq 11 S_{J+1} -210$

 Let $r_k$ be the remaining number of seats in each row.
 We have that $ r_k < s_{J+1}$ since we cannot fit the very next team into any of these rows.
 So, at least $ 11 ( 199 - S_{J+1} +1)$ seats have been taken up by students.
 Since there are $1990$ students, this means that
$ s_{J+1} + \ldots + s_I \leq 1990 - 11 ( 199 - S_{J+1} +1) = 11S_{J+1} - 210$.      

Hence, from the above, we can fit the remaining teams into the last row since
$$ s_{J+1} + \ldots + s_I  \leq 11 S_{J+1} - 210 \leq 197 < 199$$ 

Note: The extremely naive greedy algorithm (pack schools as they come), which is sufficient for the truck-stones problem, doesn't work here because we might end up with 5 teams of 39 at the end.   
We needed to apply the (still naive) greedy algorithm with the schools in order, so as to get the bound that $s_{J+1} \leq 37$ and $r_k < s_{J+1}$. This was the "more work" compared to the truck-stones problem. 
