IMO Shortlist, 2019

Let $P=\left\{i: u_{i}>0\right\}$ and $N=\left\{i: u_{i} \leqslant 0\right\}$ be the indices of positive and nonpositive elements in the sequence, and let $p=|P|$ and $n=|N|$ be the sizes of these sets; then $p+n=2019$. By the condition $\sum_{i=1}^{2019} u_{i}=0$ we have $0=\sum_{i=1}^{2019} u_{i}=\sum_{i \in P} u_{i}-\sum_{i \in N}|u_{i}|$, so
$$\begin{equation*} \sum_{i \in P} u_{i}=\sum_{i \in N}|u_{i}| . \tag{1} \end{equation*}$$
After this preparation, estimate the sum of squares of the positive and nonpositive elements as follows:
$$\begin{align*} & \sum_{i \in P} u_{i}^{2} \leqslant \sum_{i \in P} b u_{i}=b \sum_{i \in P} u_{i}=b \sum_{i \in N}|u_{i}| \leqslant b \sum_{i \in N}|a|=-n a b ; \tag{2}\\ & \sum_{i \in N} u_{i}^{2} \leqslant \sum_{i \in N}|a| \cdot|u_{i}|=|a| \sum_{i \in N}|u_{i}|=|a| \sum_{i \in P} u_{i} \leqslant|a| \sum_{i \in P} b=-p a b . \tag{3} \end{align*}$$
The sum of these estimates is
$$1=\sum_{i=1}^{2019} u_{i}^{2}=\sum_{i \in P} u_{i}^{2}+\sum_{i \in N} u_{i}^{2} \leqslant-(p+n) a b=-2019 a b ;$$
that proves $a b \leqslant \frac{-1}{2019}$.

As in the previous solution we conclude that $a<0$ and $b>0$.
For every index $i$, the number $u_{i}$ is a convex combination of $a$ and $b$, so
$$u_{i}=x_{i} a+y_{i} b \quad \text{with some weights} \ 0 \leqslant x_{i}, y_{i} \leqslant 1, \text{with} \ x_{i}+y_{i}=1 .$$
Let $X=\sum_{i=1}^{2019} x_{i}$ and $Y=\sum_{i=1}^{2019} y_{i}$. From $0=\sum_{i=1}^{2019} u_{i}=\sum_{i=1}^{2019}\left(x_{i} a+y_{i} b\right)=-|a| X+b Y$, we get
$$\begin{equation*} |a| X=b Y . \tag{4} \end{equation*}$$
From $\sum_{i=1}^{2019}\left(x_{i}+y_{i}\right)=2019$ we have
$$\begin{equation*} X+Y=2019 \tag{5} \end{equation*}$$
The system of linear equations $(4,5)$ has a unique solution:
$$X=\frac{2019 b}{|a|+b}, \quad Y=\frac{2019|a|}{|a|+b}$$
Now apply the following estimate to every $u_{i}^{2}$ in their sum:
$$u_{i}^{2}=x_{i}^{2} a^{2}+2 x_{i} y_{i} a b+y_{i}^{2} b^{2} \leqslant x_{i} a^{2}+y_{i} b^{2}$$
we obtain that
$$1=\sum_{i=1}^{2019} u_{i}^{2} \leqslant \sum_{i=1}^{2019}\left(x_{i} a^{2}+y_{i} b^{2}\right)=X a^{2}+Y b^{2}=\frac{2019 b}{|a|+b}|a|^{2}+\frac{2019|a|}{|a|+b} b^{2}=2019|a| b=-2019 a b$$
Hence, $a b \leqslant \frac{-1}{2019}$.