YES

It is convenient to use induction, although one could describe the construction explicitly. So we replace $100$ by $n$ and $199$ by $2n-1$. For $n=1$, it is trivial. Now assume that we already have a construction for $n-1$. By using our two additional rooks, we may always supply one rook on the cell B1 which serves as a substitute for the first move in the case $n-1$, applied to the lower right $(n-1) \times (n-1)$ subboard. So we apply this procedure until we have $n-1$ mutually non-attacking rooks on this board (indeed our construction will put them on the diagonal but this is irrelevant). We are thus left with $n$ rooks and the task of placing one rook on the upper left corner. Again, this can be done by induction: If we can place one rook on the cell one lower using $n-1$ rooks, we can do this once, then use the remaining $n-1$ rooks to do it again and then move one of the two rooks on this cell one up. Done. (A slightly more challenging problem would be to show that the number $2n-1$ is optimal.)