8.4. We will call a number remarkable if it can be decomposed into the sum of 2023 addends (not necessarily distinct), each of which is a natural composite number. Find the largest integer that is not remarkable.
Answer. $4 \times 2023+3=8095$.

Solution. Replace 2023 with $n$ and solve the problem in the general case for a sum of $n \geqslant 2$ composite addends. We will prove that the answer is $4 n+3$, from which we will obtain the answer to the original problem.

Claim 1. The number $4 n+3$ is not remarkable.

Proof of Claim 1. Note that $4 n+3$ is odd, so in its decomposition into a sum, there must be at least one odd addend. The smallest odd composite number is 9, and the smallest even composite number is 4. Therefore, the sum of $n$ addends is at least

$$
9+4(n-1)=4 n+5>4 n+3,
$$

from which it follows that $4 n+3$ cannot be decomposed into a sum of $n$ composite addends, that is, it is not remarkable.

Claim 2. Any natural number greater than $4 n+3$ is remarkable. Proof of Claim 2. Consider the number $a>4 n+3$. If it is even, then it can be written as

$$
a=(4 n-4)+b=b+\underbrace{4+4+\ldots+4}_{n-1},
$$

where $b \geqslant 8-$ is some even number that is composite. If $a$ is odd, then

$$
a=4 n+1+c=9+(4 n-8)+c=9+c+\underbrace{4+4+\ldots+4}_{n-2}
$$

where $c \geqslant 4-$ is some even number that is composite.

In total, the number $4 n+3$ is not remarkable, and all larger numbers are remarkable. From this, we obtain the answer to the question of the problem.

Criteria. Proof of Claim 1 or an equivalent statement - 3 points. Proof of Claim 2 or an equivalent statement - 2 points for the even and odd cases.