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. 

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. 
