Cesko-Slovacko-Poljsko, 2006

Let the numbers be $p$, $q$, $r$, $s$. Up to a permutation we may assume that $p \geq q \geq r \geq s$. We first consider the case where $p+q \geq 5$. Then
$$p^2 + q^2 + 2pq \geq 25 = 4 + (p^2 + q^2 + r^2 + s^2) \geq 4 + p^2 + q^2 + 2rs,$$
which is equivalent to $pq - rs \geq 2$.

Assume now that $p+q<5$; then
$$4 < r + s \leq p + q < 5. \tag{1}$$
Observe that
$$(pq + rs) + (pr + qs) + (ps + qr) = \frac{(p+q+r+s)^2 - (p^2+q^2+r^2+s^2)}{2} = 30.$$
Moreover,
$$pq + rs \geq pr + qs \geq ps + qr,$$
because $(p-s)(q-r) \geq 0$ and $(p-q)(r-s) \geq 0$.
We conclude that $pq + rs \geq 10$. From (1), we get $0 \leq (p+q) - (r+s) < 1$, therefore
$$(p+q)^2 - 2(p+q)(r+s) + (r+s)^2 < 1.$$
Adding this to $(p+q)^2 + 2(p+q)(r+s) + (r+s)^2 = 9^2$ gives
$$(p+q)^2 + (r+s)^2 < 41.$$
Therefore
$$41 = 21 + 2 \cdot 10 \leq (p^2 + q^2 + r^2 + s^2) + 2(pq + rs) = (p+q)^2 + (r+s)^2 < 41,$$
which is a contradiction.

Second solution.

From $a+b+c+d=9$ with ordering $a \ge b \ge c \ge d$ we have
$$\frac{a+b}{2} = \frac{9}{4} + \varepsilon_1, \quad \frac{c+d}{2} = \frac{9}{4} - \varepsilon_1$$
for some $\varepsilon_1 \ge 0$. Thus
$$a = \frac{9}{4} + \varepsilon_1 + \varepsilon_2, \quad b = \frac{9}{4} + \varepsilon_1 - \varepsilon_2, \quad c = \frac{9}{4} - \varepsilon_1 + \varepsilon_3, \quad d = \frac{9}{4} - \varepsilon_1 - \varepsilon_3$$
for some $\varepsilon_2, \varepsilon_3 \ge 0$. From $b \ge c$ we get
$$\varepsilon_1 - \varepsilon_2 \geq -\varepsilon_1 + \varepsilon_3 \quad \text{or} \quad \varepsilon_2 + \varepsilon_3 \leq 2\varepsilon_1.$$
From
$$21 = (a^2 + b^2) + (c^2 + d^2) = 2 \cdot \left(\frac{9}{4} + \varepsilon_1\right)^2 + 2\varepsilon_2^2 + 2 \cdot \left(\frac{9}{4} - \varepsilon_1\right)^2 + 2\varepsilon_3^2 = \\ = 4 \cdot \left(\frac{9}{4}\right)^2 + 4\varepsilon_1^2 + 2\varepsilon_2^2 + 2\varepsilon_3^2 = 20 + \frac{1}{4} + 2 \cdot (2\varepsilon_1^2 + \varepsilon_2^2 + \varepsilon_3^2)$$
we conclude that non-negative numbers $\varepsilon_i$ satisfy
$$2\varepsilon_1^2 + \varepsilon_2^2 + \varepsilon_3^2 = \frac{3}{8}. \tag{2}$$
Using $\varepsilon_2 + \varepsilon_3 \le 2\varepsilon_1$ and $\varepsilon_2^2 + \varepsilon_3^2 \le (\varepsilon_2 + \varepsilon_3)^2$ from (2) we obtain
$$\frac{3}{8} \leq 2\varepsilon_1^2 + (\varepsilon_2 + \varepsilon_3)^2 \leq 2\varepsilon_1^2 + 4\varepsilon_1^2 = 6\varepsilon_1^2.$$
Hence $\varepsilon_1^2 \ge (1/6) \cdot (3/8) = 1/16$ or $\varepsilon_1 \ge 1/4$. For $ab-cd$ we have
$$ab - cd = \left(\frac{9}{4} + \varepsilon_1\right)^2 - \varepsilon_2^2 - \left(\frac{9}{4} - \varepsilon_1\right)^2 + \varepsilon_3^2 = 9\varepsilon_1 - \varepsilon_2^2 + \varepsilon_3^2.$$
Substituting for $\varepsilon_2^2$ from (2) we get
$$ab - cd = 9\varepsilon_1 - \left(\frac{3}{8} - 2\varepsilon_1^2 - \varepsilon_3^2\right) + \varepsilon_3^2 = 9\varepsilon_1 + 2\varepsilon_1^2 - \frac{3}{8} + 2\varepsilon_3^2 \ge 9 \cdot \frac{1}{4} + 2 \cdot \frac{1}{16} - \frac{3}{8} = 2.$$
This ends the proof.