Solution:

We denote by $(i, j)$ the cell in the $i$-th line and in the $j$-th column for every $1 \leq i, j \leq n$. Bob associates the following pair of cells: $(i, 2k+1), (i, 2k+2)$ for $1 \leq i \leq 100$ and $0 \leq k \leq 49$ except for $(i, k) = (100, 0)$ and $(100, 1)$, and the pairs $(100, 1), (100, 3)$ and $(100, 2), (100, 4)$.
Each time Alice writes the number $j$ in one of the cells, Bob writes the number $100^{2} + 1 - j$ in the other cell of the pair.
One can prove by induction that after each of Bob's turns, for each pair of cells, either there is a number written in each of the cells of the pair, or in neither of them. And that if a number $j$ is written, $100^{2} + 1 - j$ is also written. Thus Bob can always apply the previous strategy (since $j = 100^{2} + 1 - j$ is impossible).

At the end, every line has sum $\left(100^{2} + 1\right) \times 50$.

Assume by contradiction that Alice can stop Bob from winning if he applies this strategy. Let $c_{j}$ be the sum of the numbers in the $j$-th column for $1 \leq j \leq 100$: then $c_{j} \leq 50\left(100^{2} + 1\right)$. Note that:
$$100 \times 50\left(100^{2} + 1\right) \geq c_{1} + \cdots + c_{100} = 1 + \cdots + 100^{2} = \frac{100^{2}\left(100^{2} + 1\right)}{2} = 100 \times 50\left(100^{2} + 1\right)$$
Thus we have equality in the previous inequality: $c_{1} = \cdots = c_{100} = 50\left(100^{2} + 1\right)$. But if $a$ is the number written in the cell $(100, 1)$ and $b$ the number written in the cell $(100, 2)$, then $c_{1} - b + c_{2} - c = 99\left(100^{2} + 1\right)$. Thus $b + c = 100\left(100^{2} + 1\right) - 99\left(100^{2} + 1\right) = 100^{2} + 1$: by hypothesis $c$ is also written in the cell $(100, 3)$ which is a contradiction.

Thus Bob has a winning strategy.