In town $N$ the central square hase a shape of rectangle $n \times m$, composed of squares $1 \times 1$. In order, to illuminathe the square, lanterns are placed on the corners of the tiles (including the edge of rectangle), such that every lantern illuminates all tiles in corners of which it is placed. Find the minimal amount of lanterns which can be placed, such that every tile will be illuminated even if one of the lanterns burns out.
To solve the problem of finding the minimal number of lanterns required to illuminate every tile in a \( n \times m \) rectangular grid such that every tile is illuminated by at least two lanterns, we need to consider three cases based on the parity (odd or even) of \( n \) and \( m \).

### Case 1: Both \( n \) and \( m \) are odd
1. **Coloring the Tiles**: We color the tiles such that one tile in any \( 2 \times 2 \) block of tiles is shaded. This ensures that one lantern, wherever it is placed, illuminates exactly one shaded tile.
2. **Counting Shaded Tiles**: The number of shaded tiles is \(\frac{(m+1)(n+1)}{4}\).
3. **Double Illumination Requirement**: Since every tile must be illuminated by at least 2 lanterns, the minimum number of lanterns required is \(\frac{(m+1)(n+1)}{2}\).
4. **Placement of Lanterns**: Place a red lantern at the NE corner of each shaded tile and a green lantern at the NW corner of each shaded tile. This ensures that every tile is illuminated by two lanterns.

### Case 2: One of \( n \) or \( m \) is odd, and the other is even
1. **Assumption**: Without loss of generality, assume \( m \) is odd and \( n \) is even.
2. **Coloring the Tiles**: Color the tiles as in the previous case.
3. **Counting Shaded Tiles**: The number of shaded tiles is \(\frac{(m+1)n}{4}\).
4. **Double Illumination Requirement**: The minimum number of lanterns required is \(\frac{(m+1)n}{2}\).
5. **Placement of Lanterns**: Place a red lantern at the NE corner of each shaded tile and a green lantern at the NW corner of each shaded tile. This ensures that every tile is illuminated by two lanterns.

### Case 3: Both \( n \) and \( m \) are even
1. **Perimeter Illumination**: Focus on illuminating the perimeter of the rectangle. The number of tiles in the perimeter is \( P = 2m + 2n - 4 \).
2. **Double Illumination Requirement**: To cast double-light on every tile in the perimeter, we need at least \( 2P = 4m + 4n - 8 \) units of light.
3. **Lantern Placement**: There are four places where a single lantern can cast 3 units of light on the tiles of the perimeter. All other lantern locations cast no more than 2 units of light on the tiles of the perimeter.
4. **Reduction of Light Requirement**: Claiming those 4 lantern locations reduces the number of units of light required by 12. Assuming every remaining lantern casts 2 units of light, we need no fewer than \(\frac{2P - 12}{2} = 2m + 2n - 10\) additional lanterns.
5. **Total Lanterns for Perimeter**: The total number of lanterns required to illuminate the perimeter is \( 2m + 2n - 6 \).
6. **Recursive Argument**: Apply this logic recursively to the perimeter of an \((m-4) \times (n-4)\) rectangle until \( n \) becomes 0 or 2.
7. **Final Formula**: The total number of lanterns required is \(\frac{(m+1)n}{2}\).

### Proof by Induction for Even/Even Case
1. **Base Cases**:
   - If \( n = 0 \), \( L(m, n) = \frac{1}{2}((m+1)(n)) = 0 \).
   - If \( n = 2 \), \( L(m, n) = \frac{1}{2}((m+1)(2)) = m + 1 \).
2. **Inductive Step**:
   - Assume \( L(m-4, n-4) = \frac{(m-3)(n-4)}{2} \).
   - Then, \( L(m, n) = L(m-4, n-4) + (2m + 2n - 6) \).
   - Simplifying, \( L(m, n) = \frac{(m-3)(n-4)}{2} + (2m + 2n - 6) \).
   - This simplifies to \( \frac{(m+1)n}{2} \).

Thus, in the Even/Even case, we need a minimum of \(\boxed{\frac{(m+1)n}{2}}\) lanterns.