If $n$ is odd, the answer is $\lceil \binom{n}{2}/4 \rceil$.

If $n$ is even, the answer is $\lceil \binom{n}{2}/4+n/8 \rceil$.

This follows from two special cases of a more general conjecture by Alspach.

For our purposes, we use a theorem of Heinrich, Horák, and Rosa which says that if $n \geq 7$ is odd and $a,b,c$ are such that $3a+4b+6c=\binom{n}{2}$, then $E(K_n)$ can be partitioned into $a$ $3$-cycles, $b$ $4$-cycles, and $c$ $6$-cycles. Huang and Fu proved the same result with $(3,4,6)$ replaced by $(4,5)$.

Thus, if $n \geq 7$ is odd, it is always possible to decompose $E(K_n)$ into $4$-cycles and possibly one extra cycle that is a $3$-cycle, a $5$-cycle or a $6$-cycle. The edge set of the extra cycle can obviously be covered with two $4$-cycles of $K_n$, so we are done.

If $n$ is even, then each vertex has odd degree. Let $v$ be an arbitrary vertex. Since every $4$-cycle uses $0$ or $2$ edges incident to $v$, there will be at least one edge incident to $v$ that is covered twice. Thus, in total there will be at least $n/2$ edges that are covered twice. Thus, every covering of $E(K_n)$ by $4$-cycles has size at least $\binom{n}{2}/4+n/8$. We prove that this bound can actually be achieved.

Namely, for $n$ even, Heinrich, Horák, and Rosa's result holds except with $K_n$ replaced by $K_n$ minus a perfect matching $M$, and $\binom{n}{2}$ replaced with $\frac{n(n-2)}{2}$. For $n$ even, $\frac{n(n-2)}{2}$ is divisible by $4$. It follows that the edges of $K_n-M$ can be decomposed into $4$-cycles. By then covering pairs of edges of $M$ with $4$-cycles we get a covering of size $\lceil \binom{n}{2}/4+n/8 \rceil$.

edge covering number$\rho(\mathcal{H})$ of thehypergraph$\mathcal{H}$ whose ground-set is theedge-set of the complete graph $K_n$ and whose set of hyperedges is equal to the set of edge-sets of all 4-circuits in $K_n$. Please do not be confused by the (traditional) technical term 'edge covering number': this doesnotrefer to covering theedges, rather, the term, regrettably very widespread, refers toa covering of the ground-set. $\endgroup$ – Peter Heinig Aug 29 '17 at 7:20byhyperedgesgraph-theorists call 'book covers' 'paper covers'. (I.e., the noun modifier gives not the thing-to-be-covered, rather the thing-that-the-cover-is-made-of.) $\endgroup$ – Peter Heinig Aug 29 '17 at 7:22anytwo distinct edges of the underlying graph are in a hyperedge of $\mathcal{H}$, so the Berger-Ziv-bound works out to a guarantee that there always isacover of the kind you require of size at most $\lfloor \frac{(4-2)3\binom{n}{4} + 1}{4-1}\rfloor = \lfloor \frac{n(n-1)(n-2)(n-3)}{12} + \frac13 \rfloor$. While this may be non-obvious upper bound, it's too large. E.g. for $n=5$ it gives $10$, while you gave a covering by $3$ four-circuits. $\endgroup$ – Peter Heinig Aug 29 '17 at 8:14