Solved with @Enigma714 Case a) Ana wins, consider in each column the numbers being $1$ and $8$ then the final number must be congruent to $11\dots1 \equiv \frac{10^{1001}-1}{9} \pmod{7}$ so it can't be multiple of $1001=7.13.11$ by FLT Case b) Let's see Beto wins here. Say columns are from right to left $1,2,\dots, 1003$, and in column $i$ Ana writes $1\leq a_i<b_i\leq 9$. It's clear that is enough to prove that: in a $2\times 1003$ board, where in column $i$ is written $0$ and $c_i$ (where $1\leq c_i=b_i-a_i\leq 8$ for each $i$), Beto can choose each number such that he gets an $n-$digit number (possibly with $0$s at the begging) such that he gets any possible residue $\pmod{n}$. Suppose we have that the last $k$ columns spans $t$ residues say $r_1,r_2, \dots, r_t$ we want to show that using the $k+1$ column we get at least one more residue so we are done by induction, otherwise we have the set: $$\{r_1,r_2, \dots, r_t \} = \{r_1+c_{k+1},r_2+c_{k+1}, \dots, r_t+c_{k+1} \}$$then calculate the sum of elements of each set to get $tc_{k+1} \equiv 0 \pmod{1003} $ but since $1003=17.59$ we have $(c_{k+1},1003)=1$ then $t=1003$ then we had already gotten all the remains. done