Solution:

The answer is 210. Let us consider all the 4-digit codes that can be formed with digits belonging to the correct intervals (without restrictions on the sum); these number $7 \cdot 4 \cdot 5 \cdot 3 = 420$. We now show that exactly half of these codes also satisfies the condition on the sum of the digits. We observe that if $[a, b, c, d]$ are 4 digits in the indicated intervals, then the same is true for $[6-a, 3-b, 4-c, 2-d]$; we say that the second code is the companion of the first. The sum of the digits of $[6-a, 3-b, 4-c, 2-d]$ is $15 - (a + b + c + d)$, which is greater than or equal to 8 if and only if $a + b + c + d \leq 7$; in particular, exactly one of the two codes $[a, b, c, d]$ and $[6-a, 3-b, 4-c, 2-d]$ satisfies the hypothesis on the sum of the digits. Observing furthermore that the companion of the code $[6-a, 3-b, 4-c, 2-d]$ is again $[a, b, c, d]$, we have thus formed 210 pairs, each of which contains exactly one code compatible with Eugenia's memories.