Solution:

Since, because of (2), the $r_{i}$ cannot all be $0$, and, because of (1), cannot all have the same sign, we have $b>0$ and $a<0$. With $P=\left\{i: u_{i}>0\right\}$ and $N=\left\{i: u_{i} \leq 0\right\}$ as well as $p=|P|$ and $n=|N|$ we have $p+n=2019$, and from (1) it follows that

$0=\sum_{i=1}^{2019} u_{i}=\sum_{i \in P} u_{i}-\sum_{i \in N}\left|u_{i}\right|$, hence $\sum_{i \in P} u_{i}=\sum_{i \in N}\left|u_{i}\right|$.

Thus we can estimate:

$\sum_{i \in P} u_{i}^{2} \leq \sum_{i \in P} b u_{i}=b \sum_{i \in N}\left|u_{i}\right| \leq b \sum_{i \in N}|a|=-n a b$ $\hspace{0.5cm}$ (3)

as well as

$\sum_{i \in N} u_{i}^{2} \leq \sum_{i \in N} a u_{i} \leq|a| \sum_{i \in N}\left|u_{i}\right|=|a| \sum_{i \in P} u_{i} \leq-p a b$ $\hspace{0.5cm}$ (4).

It follows that $1=\sum_{i \in P} u_{i}^{2}+\sum_{i \in N} u_{i}^{2} \leq-(p+n) a b=-2019 a b$, and thus the claim.

Solution 2:

Again starting from $b>0$ and $a<0$, we consider the following convex point set $C$ in the $x$-$y$-plane:

(i) The lower boundary of $C$ is the parabola $y=x^{2}$ in the range $a \leq x \leq b$.

(ii) The upper boundary of $C$ is the line $g: y=(a+b) x-a b$ in the range $a \leq x \leq b$.

Each of the points $(u_{i}, u_{i}^{2})$ lies on the lower boundary of $C$. Therefore the centroid $S$ of these $2019$ points, each assigned equal mass, also lies in $C$. We have $S=\left(\frac{1}{2019} \sum_{i=1}^{2019} u_{i}, \frac{1}{2019} \sum_{i=1}^{2019} u_{i}^{2}\right)=\left(0, \frac{1}{2019}\right)$. For $g$ we have at the point $x=0$ that $y=-a b$. $S$ may not lie above the upper boundary, from which the claim follows.

Solution 3:

(A one-line proof):

$0 \leq \sum_{i=1}^{2019}\left(r_{i}-a\right)\left(b-r_{i}\right)=\sum_{i=1}^{2019}\left(-r_{i}^{2}+(b+a) r_{i}-a b\right)=-1-2019 a b \Leftrightarrow a b \leq \frac{-1}{2019}$.