The answer is $2021^2$. Construction, just take $a_1 = a_2 = \cdots = a_2021 = \frac{1}{2021}$ and take $a_{2022} = 0$. Non-rigorous proof, but just sketch Now, to prove that nothing more is possible, order $a_1 \ge a_2 \ge \cdots \ge a_{2021} \ge a_{2022} \ge 0$. We want the $(a_i,a_j)$ to satisfy with $1 \le i,j \le 2021$. We also want $a_2022$ to be in atleast one pair to get the number up. Therefore, we want $(1,2022)$ to be a pair ($x^2 + y \ge y^2 + x \iff x \ge y \ge 0$). Now, we want $a_2 = a_3 = \cdots = a_{2021} = x$ to be minimal, but so that all of the pairs satisfy. Therefore, we want $2021(x^2+x) \ge 1$ or $x \ge \frac{45\sqrt{2021}-2021}{4042}$. Now, we want $a_1^2 + a_2022$ to be maximal. We have that $a_1 = \frac{45\sqrt{2021}-2021}{4042} + \epsilon$ and $a_2022 = 1 - \frac{45\sqrt{2021}-2021}{2} - \epsilon$. Clearly, this is maximal when $\epsilon$ is maximal, so $\epsilon = \frac{2023 - 45\sqrt{2021}}{2}$. However, $(a_1)^2 + (a_{2022}) = (\frac{45\sqrt{2021}-2021}{4042} + \frac{2023-45\sqrt{2021}}{2})^2 \ge \frac{1}{2021}$ is false. Therefore, $2021^2$ pairs is maximal. The only thing wrong with my proof thus far is that I assume that the number of pairs are maximized when $(i,j)$ satisfies for all $1 \le i,j \le 2021$. It could be the case that for example $(2020,2021)$ isn't a pair, but $(1,2022)$ and $(2,2022)$ are pairs increasing the number of pairs. So if you can prove that $(i,j)$ is a pair for $1 \le i,j \le 2021$ rather than $(1,2022)$ is a pair, then you're done

Very fun problem to work on, the official solution is quite nice and more natural than this solution but this is elegant nevertheless. This is exactly what I would imagine a good 2/5 at IMO to look like, $\frac{1}{2021}$ can be replaced by any positive real that is greater than $\frac{2023}{2022^2}$.

The construction, as above is going to be taking $a_1 = 0, a_2 = a_3 = \cdots = a_{2022} = \frac{1}{2021}$, I chose, reverse order as opposed to the one above by preference. Notice that if $j \in \{2, \cdots, 2022\}$, then $a_i^2+a_j \geq a_j = \frac{1}{2021}$ and there are $2021 \cdot 2022 = 4086462$ such pairs, moreover, if $a_j=1$, then $a_i^2+a_j = a_i^2 \leq (\frac{1}{2021})^2 < \frac{1}{2021}$, so indeed these are all the pairs that work; all in all we claim that the answer is $4086462$, a better way for thinking about this number being $2022^2-2022$. We may replace $2022$ by $n$, in which case we can take $a_1 = 0, a_2 = a_3 = \cdots = a_n = \frac{1}{n-1}$, with a claimed answer of $n^2-n$ (this proof does not work directly for $n=2$, in which case it is easy to check that the answer is $2^2-2 = 2$), this proof only works directly for even integers but the odd case works similarly. We wish to prove that there are at least $n$ ordered pairs $(i,j) \in [2022]^2$ such that $a_i^2+a_j < \frac{1}{2021}$, in our construction, these are all the pairs $(1,1), (2,1), \cdots, (n,1)$, and now even though this is a leap of faith, it motivates the following lemma, which immediately implies the problem. Assume without loss of generality that $a_1 \leq a_2 \leq \cdots \leq a_n$. Main Lemma: At least $n$ of the pairs $(1,1), (1,2), \cdots, (1,n), (n,1), (n-1,1), \cdots, (2,1)$ satisfy $a_i^2+a_j < \frac{1}{2021}$ $\textbf{Proof)}$ Let $k$ be such that $a_1+a_{k+1}^2, a_1+a_{k+2}^2, \cdots, a_1+a_n^2 \geq \frac{1}{2021}$ but $a_1+a_k^2, a_1+a_{k-1}^2, \cdots, a_1+a_2^2, a_1+a_1^2 < \frac{1}{2021}$. We have one Lemma that must hold for the Main Lemma to even have a chance to hold; it's therefore natural to consider if you aim to prove the Main Lemma and it turns out that it helps prove our claim. $\textbf{Lemma 1:}$ $k > \lfloor \frac{n}{2} \rfloor$ $\textbf{Proof)}$ We will only prove the lemma for the case where $n$ is even, this is the only case that matters for the problem but the other parity is not harder to check. Assume for the sake of contradiction that $k \geq \frac{n}{2}$, that is $a_1+a_n^2 \geq a_1+a_{n-1}^2 \geq a_1+a_{\frac{n}{2}}^2 \geq \frac{1}{2021}$ Notice that $a_1+a_{\frac{n}{2}} \leq \frac{1}{\frac{n}{2}}$ because $1 = \sum_{i=1}^{2022} a_i \geq \frac{n}{2} \cdot a_1 + \frac{n}{2} \cdot a_{\frac{n}{2}}$ which means that $a_{\frac{n}{2}} < \frac{1}{2}$, then $a_1+a_{\frac{n}{2}}^2 < a_1+a_{\frac{n}{2}}^2+ \varepsilon(1+\varepsilon-2a_{\frac{n}{2}}) = (a_1+\varepsilon)+(a_{\frac{n}{2}}-\varepsilon)^2$ this means that $a_1+a_{\frac{n}{2}}^2 \leq \frac{1}{2022}^2+\frac{1}{2022} < \frac{1}{n}$, a contradiction, as desired. $\blacksquare$ Now, we will proceed with the proof of the main Lemma. We can restate our goal as Lemma 2. $\textbf{Lemma 2:}$ $a_1^2+a_{n-k+1} < \frac{1}{n-1}$ $\textbf{Proof)}$ Assume for the sake of contradiction that $a_1^2+a_{n-k+1} \geq \frac{1}{2021}$. Notice that we can take $(a_k,a_{k+1}) \to (\frac{a_k+a_{k+1}}{2}, \frac{a_k+a_{k+1}}{2})$, $(a_{k+1},a_{k+2}) \to (\frac{a_{k+1}+a_{k+2}}{2}, \frac{a_{k+1}+a_{k+2}}{2}), \cdots, (a_{n-1},a_n) \to (\frac{a_{n-1}+a_n}{2}, \frac{a_{n-1}+a_n}{2})$, in the written order which keeps the conditions the same while ensuring that $a_k = a_{k+1} = \cdots = a_n = c$ for some $c \in \mathbb{R}_{\geq 0}$. The same process can be performed on the sets $\{a_1, \cdots, a_{n-k}\}$ to give $a_1 = a_2 = \cdots = a_{n-k} = a$ and $\{a_{n-k+1}, a_{n-k+2}, \cdots, a_k\}$ to give $a_{n-k+1} = a_{n-k+2} = \cdots = a_k = b$ for $a,b \in \mathbb{R}_{\geq 0}$. Now, we have the inequalities $$a^2+b \geq \frac{1}{n-1}$$$$a+c^2 \geq \frac{1}{n-1}$$where $a \leq b \leq c$ and $(n-k)a+(2k-n)b+(n-k)c = 1$. The main idea of this solution is that because $|\{a_1, a_2, \cdots, a_{n-k}\}| = |\{a_{k+1}, a_{k+2}, \cdots, a_n\}|$ if we are able to take $a_i \to a_i+\varepsilon$ for $i \in \{1, \cdots, n-k\}$, and $a_j \to a_j-\varepsilon$ for $j \in \{k+1, \cdots, n\}$ for some $\varepsilon$ small enough that $a+\varepsilon < b < c-\varepsilon$, then all the conditions are still preserved, most importantly the sum and non-negativity conditions and the fact that $(a+\varepsilon)^2+b > a^2+b \geq \frac{1}{n-1}$ and $(a+\varepsilon)+(c-\varepsilon)^2 = a+c^2 + \varepsilon(1+\varepsilon-2c) > a+c^2 \geq \frac{1}{n-1}$ being satisfied as long as $ c \leq \frac{1}{2}$, we know that $(n-k)c < 1$ meaning that if on the contrary $c > \frac{1}{2}$, then $k = n-1$. In that case $a+(n-2)b+c = 1$ with $c > \frac{1}{2}$ means that $(n-2)b \leq a+(n-2)b \leq \frac{1}{2}$ meaning that $b \leq \frac{1}{2(n-2)} \leq \frac{1}{n}$ because $n \geq 4$, therefore $$\frac{1}{n}^2+\frac{1}{n} \geq b^2+b \geq a^2+b \geq \frac{1}{n-1}$$which is a contradiction. Then, taking $\varepsilon = min(b-a, c-b)$, in the case $c \leq \frac{1}{2}$ means that we can assume that either $a=b$ or $b=c$. $\textbf{Case 1:}$ $a=b$ Then $a^2+a \geq \frac{1}{n-1}$ where $na \leq ka+(n-k)c = 1$ means that $$\frac{n+1}{n^2} = \frac{1}{n}^2+\frac{1}{n} \geq a^2+a \geq \frac{1}{n-1}$$which is a contradiction. $\textbf{Case 2:}$ $b=c$ There are many ways to deal with this case, the shortest seems to by using derivatives, we can parametrize all our variables with $k,n,b$, notice that $(n-k)a+kb = 1$ meaning that $a = \frac{1-kb}{n-k}$, therefore $a+c^2 \geq \frac{1}{n-1}$ can be rewritten as $\frac{1-kb}{n-k}+b^2 \geq \frac{1}{n-1}$ Now, let $f(x) = \frac{1}{n-k}-\frac{k}{n-k} \cdot x + x^2$ Notice that $f’(b) = 2b-\frac{k}{n-k}$ where $b < \frac{1}{2}$ because we are in the case where $c < \frac{1}{2}$ meaning that $f’(b) < \frac{1}{2}$, in fact $f’(x) < 0$ for $x \leq b$, this means that $f(b)$ is maximized when $a=b$, by noticing that $a$ will vary linearly (increase) with $b$ decreasing linearly until they meet at some point where $b = \frac{1-kb}{n-k}$, in particular in this $f(b)$ is maximized when $a=b=c$, where $\frac{n+1}{n^2} = \frac{1}{n}^2+\frac{1}{n} \geq a+c^2 \geq \frac{1}{n}$ is a contradiction. Having gotten a contradiction both cases, we can conclude $\textbf{Lemma 2}$. This means that $$\frac{1}{n-1} > a_1^2+a_{n-k+1} \geq a_1^2+a_{n-k} \geq \cdots \geq a_1+a_1^2$$and $$\frac{1}{n-1} > a_1+a_k^2 \geq a_1^2+a_{k-1}^2 \geq \cdots \geq a_1^2+a_1$$in particular there are at least $n$ pairs $(i,j)$ such that $a_i^2+a_j < \frac{1}{n-1}$ and thus at most $n^2-n$ pairs $(i,j)$ such that $a_i^2+a_j \geq \frac{1}{n-1}$, with the construction in mind this completes the proof.