Following the suggestion of Mike Shulman to check the condition for existence of the left-induced model structure:
Under the additional assumptions that $I$ be connected and that $I$-colimits commute with pull-backs, the so-called left-acyclicity condition is indeed fulfilled. I just did the exercise figuring out some assumptions we have to impose so that it becomes really easy. Presumably, they can be weakened quite considerably, but I will leave this to others.
(E.g., a very particular kind of pull-back is enough, which is more likely to commute with colimits, I guess.)

Let us call a morphism $\varphi$ in $K^I$ a cofibration (resp. a weak equivalence) if $\mathrm{colim}(\varphi)$ is a cofibration (resp. a weak equivalence) in $K$.
From Garner, Kedziorek, Riehl, Lifting accessible model structures, arXiv:1802.09889, Corollary 2.7, and the assumption that the model category $K$ is combinatorial (or more generally, accessible) it follows that we can left-induce the model structure along $\mathrm{colim}\colon K^I\to K$ if and only if the *left-acyclicity condition* holds:

*If a morphism $\varphi$ in $K^I$ satisfies the right-lifting property against all cofibrations in $K^I$, then it is a weak equivalence.*

Let $\varphi\colon A\to B$ be a morphism in $K^I$ satisfying the right-lifting property against all cofibrations in $K^I$. We have to show that it is a weak equivalence. That is, we have to show that $\mathrm{colim}(\varphi)\colon \mathrm{colim}(A)\to\mathrm{colim}(B)$ is a weak equivalence in $K$. But for this, it is enough that $\mathrm{colim}(\varphi)$ satisfies the right-lifting property agains all cofibrations in $K$.
So, let $\gamma\colon c\to d$ be a cofibration in $K$ and suppose we are given any pair of morphisms $c\to \mathrm{colim}(A)$ and $d\to \mathrm{colim}(B)$ such that the following diagram commutes.

$$\require{AMScd}\begin{CD}
c @>>> \mathrm{colim}(A)\\
@V\gamma VV @VV\mathrm{colim}(\varphi)V\\
d @>>> \mathrm{colim}(B)
\end{CD}\quad\quad(\ast)$$

I will denote the constant diagram-functor $K\to K^I$ by underlining.
Consider $C := \underline c\times_{\underline{\mathrm{colim}(A)}}A$ and $D := \underline d\times_{\underline{\mathrm{colim}(B)}}B$, where $A\to \underline{\mathrm{colim}(A)}$ and $B\to \underline{\mathrm{colim}(B)}$ are the universal morphisms.
Then we get a commutative square
$$\begin{CD}
C @>>> A\\
@V\underline{\gamma}\times\varphi VV @VV\varphi V\\
D @>>> B
\end{CD}\quad\quad(\ast\ast)$$
in $K^I$, where the horizontal morphisms are the natural projections. Now, assuming that $\mathrm{colim}(\underline c\times_{\underline{\mathrm{colim}(A)}}A) = c$, naturally in $c$ and $A$, (e.g., if $I$ is connected and $I$-colimits commute with pull-backs,) this commutative square gets mapped to $(\ast)$ under $\mathrm{colim}$. In particular, $\underline{\gamma}\times\varphi$ is a cofibration.
Thus, $\mathrm{colim}(\varphi)$ satisfies the right-lifting property agains all cofibrations as soon as $\varphi$ does, but then $\varphi$ is a weak equivalence, as claimed.

I think through a refinement of the argument, one can get rid of the connectedness-assumption. But I don't know whether this really depends on some sort of continuity.

definea map of diagrams to be a cofibration or a weak equivalence if its colimit is such. This means such a model structure would not be the projective one, and would be an answer to your question, if it exists. But you'd have to look at the conditions for such a model structure to exist and check whether they hold. $\endgroup$ – Mike Shulman Jul 20 '18 at 14:58