<s>[INST] Limit of a Combinatorial Function I need help with the following problem, proposed by Iurie Boreico:

Two players, $A$ and $B$, play the following game: $A$ divides an $n \times n $ square into strips of unit width (and various integer lengths). After that, player $B$ picks an integer $k$, $1 \leq k \leq n$, and removes all strips of length $k$. Let $l(n)$ be the largest area that $B$ can remove, regardless how $A$ divides the square into strips. Evaluate
  $$ \lim_{n \to \infty} \frac{l(n)}{n}. $$

Some progress that I made on this problem:
Observe that there is at most $l(n)/k$ strips of length $k$. So,
$l(n)n - \sum_{k = 1}^n l(n) \pmod k = \sum_{k = 1}^n k \lfloor l(n)/k \rfloor \geq n^2$.
Now I'm stumped on how to asymptotically bound the left hand side, and I cannot find this problem posted anywhere online. Any solutions, observations, progress is appreciated. 
 [/INST] Here is another way to get the bound obtained by Fedor Petrov. 
One observation is that from the covering of the board using strips of size $1 \times i$ and $1 \times n-i$ to cover row $r$ for rows $1 \leq r \leq n$, then $l(n) \leq (n-1) + (n-1) = 2n-2$. From the Pigeonhole principle, we also have that $n \leq l(n)$. So, $n \leq l(n) \leq 2n-2$. 
As stated in the OP, we have that
$$ l(n)n - \sum_{k = 1}^n l(n) \pmod k \geq n^2. $$
We will try to get a bound the second sum. To make notation more simple, we let $l(n) = L$. Splitting the sum from $1 \leq k \leq L/2$ and $L/2 < k \leq n$, we have that 
\begin{align*}
\sum_{n \geq k > L/2} l(n) \pmod k  & = \sum_{n \geq k > L/2} L - k \\
& = \left ( n - \frac{L}{2} \right ) \left ( \frac{3L}{4} - \frac{n}{2} \right) + O(L) \\
& = nL - \frac{n^2}{2} - \frac{3L^2}{8} + O(L).
\end{align*}
For the second sum, we have that
\begin{align*}
\sum_{k \leq L/2} L \pmod k & = \sum_{i = 2}^L \sum_{\frac{L}{i+1} < k \leq \frac{L}{i}} L \pmod k\\
& = \sum_{i = 2}^L \sum_{\frac{L}{i+1} < k \leq \frac{L}{i}} L - ik \\
& = \sum_{i = 2}^L \frac{L^2}{2i(i+1)^2} + L \cdot O\left (\frac{1}{i+1} \right) \\
& = O(L \log L) + \frac{L^2}{2}  \sum_{i = 2}^L \frac{1}{i(i+1)^2}.
\end{align*}
Since
$$\sum_{i = 2}^L \frac{1}{i(i+1)^2} = \sum_{i = 2}^\infty \frac{1}{i(i+1)^2} - \sum_{i \geq L+1} \frac{1}{i(i+1)^2} = \frac{7}{4} - \frac{\pi^2}{6} + O \left (\frac{1}{L^2}\right ).$$
So, 
$$\sum_{k \leq L/2} L \pmod k = O(L \log L) + \left ( \frac{7}{8} - \frac{\pi^2}{12} \right ) L^2.$$
Adding our two expressions together, we have that 
$$\sum_{k = 1}^n L \pmod k = nL - \frac{n^2}{2} + \left ( \frac{1}{2} - \frac{\pi^2}{12} \right ) L^2 + O(L \log L).$$
Hence, 
\begin{align*}
nL - \sum_{k = 1}^n l(n) \pmod k & \geq n^2 \\
\left ( \frac{\pi^2}{12} - \frac{1}{2} \right ) L^2 + O(L \log L) & \geq  \frac{n^2}{2} \\
\left ( \frac{L}{n} \right )^2 & \geq \frac{6}{\pi^2 - 6} + o(1) \\
\frac{L}{n} & \geq \sqrt{ \frac{6}{\pi^2 - 6} } + o(1).
\end{align*}
Taking the limit as $n \to \infty$, we find that the limit is $\boxed{\geq \sqrt{6/(\pi^2 - 6)}}$. 
The only upper bound I've arrived at is $2$ which can be achieved by the first observation. 
</s>