If $m\ne n$, every $N$ works. If $m=n>1$, every $N$ still works. If $m=n=1$, only $N=1$ works.

If $m\ne n$, after each player has done a move, sum of all entries increases or decreases by at least $1$ depending on $m<n$ or $m>n$. Hence, after the $mnN$-th turn of Duncan, sum of all entries is at least or at most $mnN$. Thus, one of the entries is at least or at most $N$. Thus, Bruce will win in this case. Now, suppose $m=n>1$, then Bruce just have to keep picking the same row every turn. Since $n>1$ by assumption, each of Bruce's moves increases the sum of the picked row by at least $2$ while Duncan can only decrease it by $1$. Thus, after $nN$ moves of Bruce, one of the entries is at least $N$. The case for $m=n=1$ is obvious. The only entry on the board will be always be $0$ or $1$ alternatively. Hence, Bruce wins if and only if $N=1$.