Related problem: https://artofproblemsolving.com/community/c6h597127p3543383

Just for convenience of the reader, let me put a solution to this problem here:

A solution exploiting $2n$ is even (which unfortunately doesn't work if the grid was $\text{odd} \times \text{odd}$): The answer is $2k-1$. The construction is to put numbers $\frac{k}{2k-1}$ is some $2k-1$ cells and $0$ is all the other cells (here we require $2k - 1 \le 4n^2$). Note that any piece must contain at most one non-zero number, and hence there are at least $2k-1$ pieces. Now we prove that $2k-1$ pieces always suffice. Claim (Optimization) : We may Wlog assume that if sum of numbers in some piece is $\le 1$, then it contains at most one non-zero number. Proof: Otherwise, we iterate the numbers in the grid a bit: we put sum of all numbers of that piece in some particular cell of that and put $0$ at other places. It is not hard to note that if we can solve this new problem, then the old problem can also be solved. $\square$ Now denote by $f(i)$ the number of non-zero cells in the $i$-th row and $s(i)$ the sum of numbers in the $i$-th row. Claim: For any $1 \le i \le 2n-1$, we have $s(i) + s(i+1) > \frac{f(i) + f(i+1)}{2}$. Proof: Fix any $i \in \{1,2,\ldots,2n-1\}$. Let $a_1,a_2,\ldots,a_k ; b_1,b_2,\ldots,b_l$ be the non-zero numbers of the $i$-th , $(i+1)$-th row (respectively), from left to right. Note that there is a connnected piece containing the only non-zero numbers $a_j$ and $a_{j+1}$ for any $1 \le j \le k-1$. $b_j$ and $b_{j+1}$ for any $1 \le j \le l-1$. $a_1$ and $b_1$. $a_k,b_l$. Hence we obtain \begin{align*} 2 \Big( s(i) + s(i+1) \Big) &= 2 \Big( (a_1 + a_2 + \cdots a_k) + (b_1 + b_2 + \ldots + b_l) \Big) \\ &= \sum_{j=1}^{k-1} (a_j + a_{j+1}) + \sum_{j=1}^{l-1} (b_j + b_{j+1}) + (a_1 + b_1) + (a_k + b_l) \\ &> \sum_{j=1}^{k-1} 1 + \sum_{j=1}^{l-1} 1 + 1 + 1 = (k-1) + (l-1) + 1 + 1 = k+l = f(i) + f(i+1) \end{align*}This proves our claim. $\square$ Hence we obtain \begin{align*} k &= \Big( s(1) + s(2) \Big) + \Big(s(3) + s(4) \Big) + \cdots + \Big(s(2n-1) + s(2n) \Big) \\ &> \left( \frac{f(1) + f(2)}{2} \right) + \left( \frac{f(3) + f(4)}{2} \right) + \cdots + \left( \frac{f(2n-1) + f(2n)}{2} \right) \\ & = \frac{f(1) + f(2) + \cdots f(2n)}{2} = \frac{\# \text{ non-zero numbers in the table}}{2} \end{align*}It follows that $\# \text{ non-zero numbers in the table}$ is at most $2k-1$. So clearly $2k-1$ pieces suffice. This completes the proof of the problem. $\blacksquare$

My solution. $2k-1$ is the solution. Note that Lee can always choose to put $\frac{k}{2k-1}$ on $2k-1$ grids, that way, sunny must divide the paper into at least $2k-1$ pieces, as if two of these grids are in one piece then the sum of which is at least $\frac{2k}{2k-1} > 1$. We now claim that no matter $n, k$, Lee cannot force Sunny to divide the paper into more than $2k-1$ pieces. Choose a connected path that visits all grid- we can further require that the last visited grid connects to the first. For any setup Lee gives, we perform the following algorithm: We keep visiting new grids until the sum of numbers on the grids exceeds $1$, then we cut the visited grids into a full piece of paper, then perform the same steps until we visit all grids. We claim the algorithm terminates before cutting $2k$ pieces, which finishes the problem. If we are left with more than $2k$ pieces, then consider the sum of the grids on the odd # of pieces that was cut, in addition to all the next grids of which the algorithm detects that the sum along with the piece exceeds $1$. This sum is greater than $1 \times k = k$, with all grids counted as most once, a contradiction.