Notice first of all that if $|L|$ has a fixed part, then applying Clifford's theorem to the moving part of $|L|$ one obtains immediately that the fixed part is a point, $C$ is hyperelliptic and $|L|$ is a multiple of the $g^1_2$.
So we can assume that $|L|$ is free.

Another remark is that $K_C-L$ also satisfies $h^0(K_C-L)=\deg(K_C-L)/2$, so we can also assume $\deg L\le g(C)-1$.

There exists a more precise form of Clifford's theorem, sometimes called "Clifford+" that can be useful for answering your question. The statement is the following (cf. [Beauville, "L'application canonique pour les surfaces de type g\'en\eral", Invent. Math. 55, 121-140, (1979], Lemma 5.1):

Let $C$ be a smooth curve and $L$ a divisor of $C$ with $\deg L\le g(C)-1$. Then one of the following holds:

(i) $h^0(L)\le \deg L/3+4$;

(ii) there exists a a degree 2 map $f\colon C\to \Gamma$ such that $\Gamma$ is a smooth curve and the moving part of $|L|$ is equal to $f^*|\Delta|$ where $\Delta$ is a divisor of $\Gamma$. In this case $h^0(L)\le \deg L/2+1-g(\Gamma)$.

In your case you get immediately that for $h^0(L)\ge 13$ the map given by $|L|$ factors through a double cover $C\to \Gamma$, where $\Gamma$ is either rational or elliptic. If $\Gamma$ is rational, then $C$ is hyperelliptic and $L$ is multiple of the $g^1_2$, but this is not possible since we are assuming that $2h^0(L)=\deg L$. So $C$ is bielliptic and $|L|$ is pulled back from an elliptic curve.

The cases with $r:=h^0(L)-1\le 11$ I think should be done by hand. For $r=1$ any tetragonal curve will do, as pointed out in the comments. For $r>1$, the image $C'$ of $C$ via $|L|$ has degree $\ge r$, hence if $r>2$ then the map given by $|L|$ has degree $\le 2$. If the degree is 2 then $C'$ has degree $r+1$ hence $C'$ is elliptic (it cannot be rational, otherwise the system $|L|$ would not be complete) and we are in the previous situation. If the map given by $|L|$ is birational, then you can use Castelnuovo's bound to give an upper bound for $g(C)$ in terms of $r$. In case $r=2$ there's one more possibility: $C'$ is a conic and $|L|$ gives a map of degree 3. This corresponds to the case in which $C$ is a trigonal curve and $L$ is twice a $g^1_3$.