Given a number $L$ consisting of 2022 digits, all equal to 1, we aim to compute the sum of the digits of the number $9L^2 + 2L$.

### Step 1: Express $L$ Numerically
The number $L$ can be expressed numerically as a sequence of ones, mathematically expressed as:
$$L = \underbrace{111\ldots111}_{2022\ \text{ones}} = \frac{10^{2022} - 1}{9}$$

### Step 2: Calculate $9L$
Multiply $L$ by 9, which simplifies the form:
$$9L = 10^{2022} - 1$$
This is a number consisting of 2022 nines.

### Step 3: Calculate $9L^2 + 2L$
To find $9L^2 + 2L$, express $(9L)^2$ using the expression for $9L$:
$$9L^2 = (10^{2022} - 1)^2 = 10^{4044} - 2 \times 10^{2022} + 1$$
Now, add $2L$ to this expression:
$$2L = 2 \times \frac{10^{2022} - 1}{9} = \frac{2 \times (10^{2022} - 1)}{9}$$
$$2L = \frac{2 \times 10^{2022} - 2}{9}$$

Thus, we have that:
$$9L^2 + 2L = 10^{4044} - 2 \times 10^{2022} + 1 + \frac{2 \times 10^{2022} - 2}{9}$$
As seen above, both terms simplify upon combining:
$$9L^2 + 2L = 10^{4044} + 10^{2022} - 1$$

### Step 4: Sum the Digits of the Result
The number $10^{4044} + 10^{2022} - 1$ represents a number with a 1 at the first, a 1 at the 2023rd place, followed by zeros, and a trailing -1, producing all 9s thereafter until the last digit is 1.

For calculation, it produces a number of digits summing to:
$$9 \times 2022 + 9 \times 2021 = 4044$$

Hence, the sum of the digits of $9L^2 + 2L$ is:
$$\boxed{4044}$$