Very nice problem. We claim that the only solution is $f(x)=x$, which clearly works. Let $A = \{ a \ge 0 \colon \exists \, b \in \mathbb{R} \colon \forall \, x \in \mathbb{N} \colon f(x) \ge ax + b \}$. First note that $P(1,x)$ implies for all $x \in \mathbb{N}$ that $$f(x) > f(1) + xf(f(1)) - 2022 \ge x - 2022 \Longrightarrow 1 \in A.$$Moreover, $P(x,1)$ implies for all $x \in \mathbb{N}$ that $f(x) < (1+f(1))x + 2021$. This shows that $a \le 1+f(1)$ for every $a \in A$. Next, assume that $a \in A$ and let $b \in \mathbb{R}$ satisfy for all $x \in \mathbb{N}$ that $f(x) \ge ax+b$. Then, we have for all $x,y \in \mathbb{N}$ that $$xf(y) > ax+b - 2021 - x + (a(ax+b)+b)y \Longrightarrow f(y) > a^2y + a-1 + \frac{b-2021+ (ab+b)y}{x}.$$Taking $x \to \infty$ yields $f(y) \ge a^2y+a-1$. Hence, $a^2 \in A$. Now if $a$ contains any element $>1$, by applying this argument repeatedly we obtain that $A$ is unbounded, which is a contradiction. Hence, $1$ is the maximal element of $A$. Since $f(y) > \frac{f(f(x))}{x} y + \frac{f(x)-2021}{x}-1$ for all $x,y$, we must have $\frac{f(f(x))}{x} \in A$ and, consequently, $f(f(x)) \le x$ for all $x$. Combining this with $f(x) \ge x-2022$ yields $f(x) \le f(f(x))+2022 \le x+2022$. Hence, $\lim_{x \to \infty} \frac{f(x)}{x} = \lim_{x \to \infty} \frac{f(f(x))}{x} = 1$. Finally, $P(x,y)$ implies $$f(y) > \frac{f(f(x))}{x} y + \frac{f(x)}{x} -1- \frac{2021}{x},$$and with $x \to \infty$ we get $f(y) \ge y$ for all $y$. Together with $f(f(x)) \le x$ this yields $f(x)=x$, as desired.

hakN wrote: Find all functions $f : \mathbb{N} \to \mathbb{N}$ with $$f(x) + yf(f(x)) < x(1 + f(y)) + 2021$$holds for all positive integers $x,y.$ Let $P(x,y)$ be the assertion of the problem. By $P(1,y)$ we have $f(y)> y-a$ where $a= 2022$. Then we have $$xf(y)+x+a> x-a + y(f(x)-a)> xy+x -(a+2ay) \rightarrow f(y) +\frac{a}{x} > y-\dfrac{a+2ay}{x}. $$Letting $x\rightarrow +\infty$ in the last inequality we conclude that for all $y\in \mathbb N$ : $f(y)\geq y$. By $P(x,x)$ we have $$a+ x+xf(x) > f(x)+xf(f(x)) \rightarrow \dfrac{a}{xf(x)} + \dfrac{x-1}{x} > \dfrac{f(f(x))-1}{f(x)}$$Here, $\operatorname{LHS}$ is less than $1-\dfrac{x-a}{x^2}$ which is less than $1$ for large enough $x$. So there is $N>0$ such that for all $x>N$ we have $f(x) >f(f(x))-1$ and because the range of $f$ in $\mathbb N$ we conclude that $$\forall x>N:\;\; f(x)\geq f(f(x)),$$for which besides the fact that $f(n)\geq n$ for any $n$, we get that $f(f(x)) = f(x)$ for all $x>N$. Therefore, for $y>N$ in $P(x,f(y))$ we have $$a+x(1+f(f(y))) = a+x(1+f(y)) >f(y)f(f(x))+f(x) \geq f(x)(1+f(y)) \rightarrow \dfrac{a}{1+f(y)} > f(x)-x.$$Then by tending $y$ to $+\infty$ we get that $f(x)\leq x$ for all $x\in \mathbb N$ and then we get $f(x)=x$ for all natural numbers $x$.

$$f(x) + yf(f(x)) \leq x(1 + f(y)) + 2020$$Answer is $f(x)=x$ for all positive integers. Claim: $-2020\leq x-f(x)\leq 2020$. Proof: $P(1,y)\implies 2021+f(y)\geq f(1)+yf(f(1))\geq 1+y\implies 2020\geq y-f(y)$ $P(x,f(x))\implies f(x)+f(x)f(f(x))\leq x+xf(f(x))+2020\iff (f(x)-x)(1+f(f(x)))\leq 2020$ If $x\geq 6060,$ then $f(f(x))+1\geq f(x)-2019\geq x-4039\geq 2021$ thus, $f(x)\leq x$ for sufficiently large $x$. \[x+2020\geq f(x)+yf(f(x))-xf(y)\geq x-2020+yf(f(x))-xf(y)\]Take $y$ sufficiently large to get \[4040\geq yf(f(x-xf(y)\geq yf(f(x))-xy=y(f(f(x))-x)\]Hence $f(f(x))\leq x$. Note that this implies $f(f(1))=1$. \[P(x,f(1)): \ 2020+2x=2020+x+xf(f(1))\geq f(x)+f(f(x))f(1)\geq f(x)+f(f(x))\geq 2f(x)-2020\]Thus, $f(x)\leq x+2020\iff x-f(x)\geq -2020$.$\square$ \[x+xf(y)+2020\geq f(x)+yf(f(x))\geq f(x)+y(f(x)-2020)\]\[yf(x)-xf(y)\leq x+2020-f(x)+2020y\leq 4040+2020y=2020(y+2)\]If we have $f(y)<y$ for a sufficiently large $y$, then \[2020(y+2)\geq yf(x)-xf(y)\geq yf(x)-x(y-1)=y(f(x)-x)+x\geq x-2020y\]But this is impossible for $x>4040y+4040$. Hence $f(x)=x$ for all sufficiently large $x$. Take $x>2020^{2020}$ such that $f(x)=x,f(f(x))=x$. Then, \[x+xf(y)+2020\geq x+xy\iff 2020\geq x(y-f(y))\]Hence $y\leq f(y)$ for all $y\in Z^+$. Now we get that \[x+xf(y)+2020\geq f(x)+yf(f(x))\geq f(x)+yf(x)\geq x+yf(x)\]Thus, $2020\geq yf(x)-xf(y)$. Choose $y$ sufficiently large to verify $2020\geq yf(x)-xy=y(f(x)-x)$ which yields $f(x)\leq x$. So we have $f(x)=x$ for all $x\in Z^+$ as desired.$\blacksquare$