Answer: $10003$.
We need to find the shortest sequence of decimal digits that includes every 4-digit sequence as a (consecutive) subsequence. Since there are $10^4$ possible codes, and an $n$-digit sequence has at most $n-3$ distinct 4-digit subsequences, the answer must be at least $10^4 + 3 = 10003$.

We will construct a sequence of length $10^4+3$ that fulfills the condition. We define a directed graph $G = (V, E)$ whose vertex set $V$ consists of all 3-digit sequences. The edge set $E$ contains all ordered pairs of the form $(abc, bcd)$. (This implies that also pairs $(aaa, aaa)$ are edges.) The in-degree of a vertex is defined as the number of edges directed toward the vertex, obviously, the in-degree of every vertex in $G$ is $10$. Symmetrically, the out-degree of every vertex is also $10$. Note also that $G$ is (strongly) connected.

By a well-known argument, a connected directed graph has an Euler circuit if and only if the in-degree of any vertex equals its out-degree. We can now construct the required sequence as follows. Start from any vertex, say $000$, and begin the sequence by the digits of that vertex. Follow the Euler circuit, and at each edge $(abc, bcd)$ add $d$ to the sequence. It is easily proved by induction that after the edge $(abc, bcd)$ has been followed, the four last digits of the sequence are $ABCD$. Since the Euler circuit passes through every edge, the resulting sequence includes every 4-digit sequence. There are $10^3$ vertices, each of which is the starting point of $10$ edges, so all in all there are $10^4$ edges. Thus the length of the sequence is $3 + 10^4$ as required.