Taiwan IMO Selection Camp, 2021

The answer is $2k-1$. Let us consider the strategies of the two players in order.

- For Lee, he can write $\frac{1}{2}+\epsilon$ in $2k-1$ cells, write $\frac{1}{2}-(2k-1)\epsilon$ in one cell, and fill the rest with 0. Since each piece can contain at most one cell with $\frac{1}{2}+\epsilon$ written in it, Lee can use this strategy to guarantee $M \ge 2k-1$.

- For Sunny, he first chooses a path that passes through all the cells in one stroke, and suppose the numbers on this path, in order, are $a_1, a_2, \dots, a_{4n^2}$. Sunny's strategy is as follows: starting from $a_1$, find the largest positive integer $M$ such that $\sum_{i=1}^M a_i \le 1$, and cut $a_1$ through $a_M$ into one piece. Then, starting from $a_{M+1}$, repeat the same action, until finished.
Let us prove that the above strategy will cut out at most $2k-1$ pieces. Suppose that in the end $t \ge 2k$ pieces are cut out, and let the sum of the numbers on each piece be $b_1, b_2, \dots, b_t$. By the strategy, we know that $b_i + b_{i+1} > 1$ (otherwise they would be placed in the same piece). But then,
$$k = \sum_{t=1}^{k} b_i \ge \sum_{i=1}^{k} (b_{2i-1} + b_{2i}) > k,$$
a contradiction! Hence Sunny can guarantee $M \le 2k-1$.