Divide the numbers in $3$ sets of the form $3k,3k+1$ and $3k+2$
$S_1=\left\{1,4,7,\dots \dots ,88\right\}$
$S_2=\left\{2,5,8,\dots \dots .,89\right\}$
$S_3=\left\{\mathrm{3,6},9,\dots \dots ,90\right\}$
For the sum of two numbers to be divisible by three, either we will choose the two numbers from $S_3$ or one number from $S_1$ and one from $S_2$

i.e. $k=\left({}_{}{}^{30}C_1\times {}_{}{}^{30}C_1\right)+{}_{}{}^{30}C_2=30\times 30+\frac{30\times 29}{2}=1335$

Hence, $\frac{1335}{500}=2.67$