A number of $n > m \geq 1$ soccer teams play a full tournament, each team meeting (once) each other. Points are awarded: $2$ for a victory, $1$ for a tie and $0$ for a loss. At the end, each team has won half of its points against the $m$ teams placed last (including each of these teams, who won half of its points against the other $m-1$).

Find all possible values for $n$ and $m$, supported with examples of such tournaments.
1. **Define the teams and points:**
   Let us call the last \( m \) teams the "losers" and the other \( n-m \) teams the "winners". Each team plays against every other team exactly once. Points are awarded as follows: \( 2 \) points for a victory, \( 1 \) point for a tie, and \( 0 \) points for a loss.

2. **Calculate total points:**
   The total number of matches played in the tournament is \( \binom{n}{2} = \frac{n(n-1)}{2} \). Since each match awards \( 2 \) points in total (either \( 2 \) points to the winner or \( 1 \) point to each team in case of a tie), the total number of points awarded in the tournament is:
   \[
   2 \cdot \binom{n}{2} = 2 \cdot \frac{n(n-1)}{2} = n(n-1)
   \]

3. **Points won by losers and winners:**
   According to the problem, the losers together have won \( 4 \cdot \binom{m}{2} \) points, and the winners together have won \( 4 \cdot \binom{n-m}{2} \) points. Therefore, we have:
   \[
   4 \cdot \binom{m}{2} + 4 \cdot \binom{n-m}{2} = n(n-1)
   \]

4. **Simplify the equation:**
   \[
   4 \cdot \frac{m(m-1)}{2} + 4 \cdot \frac{(n-m)(n-m-1)}{2} = n(n-1)
   \]
   \[
   2m(m-1) + 2(n-m)(n-m-1) = n(n-1)
   \]
   \[
   2m^2 - 2m + 2(n^2 - 2mn + m^2 - n + m) = n^2 - n
   \]
   \[
   2m^2 - 2m + 2n^2 - 4mn + 2m^2 - 2n + 2m = n^2 - n
   \]
   \[
   4m^2 + 2n^2 - 4mn - 2n = n^2 - n
   \]
   \[
   2n^2 - 4mn + 4m^2 - 2n = n^2 - n
   \]
   \[
   n^2 - 4mn + 4m^2 - n = 0
   \]
   \[
   n^2 - (4m+1)n + 4m^2 = 0
   \]

5. **Solve the quadratic equation:**
   The quadratic equation \( n^2 - (4m+1)n + 4m^2 = 0 \) can be solved using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   n = \frac{4m+1 \pm \sqrt{(4m+1)^2 - 16m^2}}{2}
   \]
   \[
   n = \frac{4m+1 \pm \sqrt{16m^2 + 8m + 1 - 16m^2}}{2}
   \]
   \[
   n = \frac{4m+1 \pm \sqrt{8m + 1}}{2}
   \]
   For \( n \) to be an integer, \( 8m + 1 \) must be a perfect square. Let \( 8m + 1 = k^2 \) for some integer \( k \):
   \[
   8m = k^2 - 1
   \]
   \[
   m = \frac{k^2 - 1}{8}
   \]
   Since \( m \) must be an integer, \( k^2 - 1 \) must be divisible by \( 8 \). This implies \( k \) must be odd. Let \( k = 2j + 1 \) for some integer \( j \):
   \[
   m = \frac{(2j+1)^2 - 1}{8} = \frac{4j^2 + 4j + 1 - 1}{8} = \frac{4j(j+1)}{8} = \frac{j(j+1)}{2}
   \]

6. **Determine \( n \):**
   \[
   n = \frac{4m+1 + k}{2} = \frac{4 \cdot \frac{j(j+1)}{2} + 1 + (2j+1)}{2} = \frac{2j(j+1) + 1 + 2j + 1}{2} = \frac{2j^2 + 2j + 2j + 2}{2} = j^2 + 2j + 1 = (j+1)^2
   \]

7. **Verify the conditions:**
   We need to check that \( n \geq 2m \):
   \[
   (j+1)^2 \geq 2 \cdot \frac{j(j+1)}{2}
   \]
   \[
   (j+1)^2 \geq j(j+1)
   \]
   \[
   j^2 + 2j + 1 \geq j^2 + j
   \]
   \[
   2j + 1 \geq j
   \]
   This is always true for \( j \geq 1 \).

Thus, the possible values for \( n \) and \( m \) are:
\[
n = (k+1)^2 \quad \text{and} \quad m = \frac{k(k+1)}{2}
\]
where \( k \geq 1 \).

The final answer is \( \boxed{ n = (k+1)^2 } \) and \( m = \frac{k(k+1)}{2} \) for \( k \geq 1 \).