Yes, 23 has an inverse mod 1000 because gcd(23, 1000) = 1 (i.e. they are coprime).
Let x be the inverse. Then x is such that
23x ≡ 1 (mod 1000)
Use the Euclidean algorithm to solve for x :
1000 = 43×23 + 11
23 = 2×11 + 1
→ 1 ≡ 23 - 2×11 (mod 1000)
→ 1 ≡ 23 - 2×(1000 - 43×23) (mod 1000)
→ 1 ≡ 23 - 2×1000 + 86×23 (mod 1000)
→ 1 ≡ 87×23 - 2×1000 ≡ 87×23 (mod 1000)
→ 23⁻¹ ≡ 87 (mod 1000)
