To solve the problem, we first observe the pattern in the remainders:

Step 1: Identify the relationship between (n) and divrs

For each divr (d \in {3,4,5,6}), (n \equiv d-2 \mod d), meaning (n+2) is divisible by all these divrs. Thus, (n+2) is the least common multiple (LCM) of 3,4,5,6.

Step 2: Compute LCM of 3,4,5,6

Prime factors: (3=3), (4=2^2), (5=5), (6=2×3).
LCM = (2^2×3×5 = 60).

Thus, (n+2=60k) → (n=60k-2) (where (k) is a positive integer).

Step 3: Find valid (k) for (1 \leq n \leq 2023)

(60k-2 \leq 2023) → (60k ≤ 2025) → (k ≤ 33.75).
(60k-2 ≥1) → (k≥1) (since (k=0) gives negative (n)).

Valid (k) values: 1 to 33 → total of 33 terms.

Answer: (\boxed{33})

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