Belarusian Mathematical Olympiad

Answer: $2(n+1)\lfloor \frac{m+1}{2}\rfloor$, where $\lfloor x\rfloor$ is the greatest integer not exceeding $x$.

We paint some tiles of the town square black (if $m=2k+1$ is odd, then see Fig. 1, if $m=2k$ is even, then see Fig. 2).

It is easy to see that any lamp can illuminate at most one painted tile. By condition, any tile must be illuminated by at least two lamps. It follows that the minimum number of the lampposts is greater than or equal to $2k$, where $l$ is the number of painted tiles. If the length of the square is odd, i.e. $m=2k+1$, where $k\ge 0$, then $l=(n+1)(k+1)=(n+1)\lfloor \frac{m+1}{2}\rfloor$; if $m=2k$, then $l=(n+1)k=(n+1)\lfloor \frac{m+1}{2}\rfloor$.

On the other hand, if we place the lampposts as it is shown in the figures (the lampposts are indicated as uncolored circles), then any of the square tiles will be illuminated by two lamps. So, $2l=2(n+1)\left[\frac{m+1}{2}\right]$ lampposts are sufficient to illuminate the town square so that the problem condition will hold.