Without the floor the sum is $\frac{1}{5}\bigl(\sum n^2 - a\sum n\bigr);$ there is an integer $a$ making it exactly $0.$

For that $a,$ $U = -\frac{1}{5}\sum\bigl((n^2 - na) \bmod 5\bigr),$ and the residues repeat with period $5.$

Ignoring the floors, $\sum_{n=1}^{2023} \frac{n^2 - na}{5}$ $= \frac{1}{5}\left(\sum n^2 - a\sum n\right)$ vanishes exactly when $a = \frac{\sum n^2}{\sum n} = \frac{2 \cdot 2023 + 1}{3} = 1349,$ an integer. For any other integer $a$ the raw sum has absolute value at least $\frac{1}{5}\sum n = \frac{2023 \cdot 1012}{5} \approx 409455,$ while taking floors changes the total by less than $2023,$ so only $a = 1349$ can put $U$ strictly between $-1000$ and $1000.$

With $a = 1349,$ each term is $\frac{n^2 - 1349n - r_n}{5}$ with $r_n = (n^2 - 1349n) \bmod 5,$ so $U = -\frac{1}{5}\sum r_n.$ Since $1349 \equiv 4 \pmod 5,$ we have $n^2 - 1349n$ $\equiv n(n+1) \pmod 5,$ whose residues for $n \equiv 0, 1, 2, 3, 4$ are $0, 2, 1, 2, 0,$ summing to $5$ per block of five. With $2023 = 5 \cdot 404 + 3,$ the leftover terms $n \equiv 1, 2, 3$ contribute $2 + 1 + 2 = 5,$ so $\sum r_n = 405 \cdot 5 = 2025.$

So $U = -\frac{2025}{5} = -405,$ which indeed lies strictly between $-1000$ and $1000,$ and $a + U = 1349 - 405 = 944.$