IRL ABooklet

Let the four integers be $n_1, n_2, n_3, n_4$ and write them as
$$n_j = 3x_j - r_j \quad x_j \ge 1, r_j = 1 \text{ or } 2.$$
Without any further constraint, there are $16$ possible combinations of values $1$ and $2$ for the $r_j$. However, as $2021 \equiv 2 \pmod 3$, we must have
$$r_1 + r_2 + r_3 + r_4 \equiv -2 \equiv 1 \pmod 3,$$
and given the range of the $r_j$ this implies $r_1 + r_2 + r_3 + r_4 = 4$ or $7$. The first of these has one case, the second has four:

Let $s = x_1+x_2+x_3+x_4$, then $2021 = n_1+n_2+n_3+n_4 = 3s+(r_1+r_2+r_3+r_4)$ and
$$s = \frac{2021 + r_1 + r_2 + r_3 + r_4}{3} = 675 \text{ or } 676.$$
The number of solutions to the original problem is equal to the number of solutions to $x_1+x_2+x_3+x_4 = 675$ plus four times the number of solutions to $x_1+x_2+x_3+x_4 = 676$. By considering the numbers $\{x_1, x_1+x_2, x_1+x_2+x_3\}$ as a subset of $\{1, 2, 3, \dots, s-1\}$ we conclude that the number of solutions $(x_1, x_2, x_3, x_4)$ to $x_1 + x_2 + x_3 + x_4 = s$ is
$$\binom{s-1}{3}.$$
Therefore, the number of solutions to the original problem is
$$$$
$\binom{674}{3} + 4\binom{675}{3}$ = 674 $\cdot 673 \cdot$ 562 = 254\,924\,324.