I look for a reference for the following problem. Given an integer $k$, find a set $A\subset\mathbb{N}$ with $A=k$ that maximizes $t$ such that $\left[0,1,..,t\right]\subset A+A$.

$\begingroup$ For low numbers:$$k=1, t=0: \{0\}$$ $$k=2, t=2: \{0,1\}$$ $$k=3, t=4: \{0,1,2\} \text{ or } \{0,1,3\}$$ $$k=4, t=8: \{0,1,3,4\}$$ $\endgroup$ – Matt F. Aug 8 at 19:34

2$\begingroup$ $$k=5, t=12: \{0,1,3,5,6\}$$Also, Oeis.org/A126684 can be used to find lower bounds for $t$. However, none of its OEIS crossreferences begin with $0,2,4,8,12$, and none of the OEIS sequences beginning $0,2,4,8,12$ look promising  so existing literature may have little to say on the sequence in the question. $\endgroup$ – Matt F. Aug 8 at 20:32
A table of values for these $t$ are given in the introduction Graham and Sloane's On Additive Bases and Harmonius Graphs (your sequence corresponds to $n_\beta(k)$ in their notation). Graham and Sloane also give some references to previous work with this sequence, both under the name of "interval basis" (or Abschnittsbasis), going back to a paper in German from Rohrbach in the 1930's, and under the name of "The Postage Stamp Problem".
This is sequence A001212 in the OEIS, which has additional references.

$\begingroup$ Glad you found this or knew it! Now I see why I missed it...I got confused by not seeing 0's and thinking that the postage stamp problem was about twodimensional configurations of stamps instead. $\endgroup$ – Matt F. Aug 8 at 21:39
This is related to ``thin additive bases" of order $2$. Clearly $t$ cannot be larger than $k(k+1)/2$. It is also possible to give examples where $t$ grows quadratically. Take $A=A_0 \cup A_1$ where $A_0$ contains all integers below $t$ with binary expansion $\sum_{j} \epsilon_j 2^j$ with $\epsilon_j= 0$ unless $j$ is even, and $A_1$ consists of numbers with binary digits $\epsilon_j=0$ unless $j$ is odd. Then $A$ has $O(\sqrt{t})$ elements in it; or alternatively $t\ge Ck^2$ for some constant $C>0$. See for example this paper of Blomer which has other references.

4$\begingroup$ or simply take $A=\{0,1,\ldots,m1\}\cup \{m,2m,3m,\ldots,m^2\}$ for $m=\lfloor k/2 \rfloor$ $\endgroup$ – Fedor Petrov Aug 8 at 20:59