Bulgarian National Olympiad

We show that the desired number equals $t = 2k - 2$.

Consider the set $S = \{3, 4, \dots, 2k, 2k+1\}$. The sum of any two distinct numbers from $1, 2, \dots, k+1$ is an element of $S$. Since among $1, 2, \dots, k+1$ there exist two numbers having one and the same color we conclude that $S$ is not good. Now $|S| = 2k-1$ implies $t \le 2k-2$.

It remains to prove that the set $S = \{a+1, a+2, \dots, a+2k-2\}$ is good for any $a$.

1. Let $a$ be an odd number. Color the numbers $1, 2, \dots, \frac{a+1}{2}$ in the first color and every of the numbers $\frac{a+2s-1}{2}$ for $s = 2, 3, \dots, k$ in color $s$. Let all numbers greater than $\frac{a+2k-1}{2}$ be of color $k$. It is easy to see that the sum of any two numbers of one and the same color is not an element of $S$.

2. Let $a$ be an even number. Color the numbers $1, 2, \dots, \frac{a}{2}$ in the first color and every of the numbers $\frac{a+2s-2}{2}$ for $s = 2, 3, \dots, k$ in color $s$. Let all numbers greater than $\frac{a+2k-2}{2}$ be of color $k$. It is easy to see that the sum of any two numbers of one and the same color is not an element of $S$.