36. Subtract from the sum of all positive three-digit numbers $S^{(1)}$ the sum of positive three-digit numbers divisible by 2 ($S^{(2)}$), and the sum of positive three-digit numbers divisible by 3 ($S^{(3)}$). In this process, numbers divisible by both 2 and 3, i.e., divisible by 6, are subtracted twice.

Therefore, to get the answer, we need to add the sum of all positive three-digit numbers divisible by 6 ($S^{(6)}$) to the obtained difference, i.e., take the value

$$
S=S^{(1)}-S^{(2)}-S^{(3)}+S^{(6)}
$$

Each of the four sums written on the right is the sum of a certain number of terms of the corresponding arithmetic progression. The number of terms and the common difference of the progression are given in the table.

|  | $u_{1}$ | $n$ | $d$ |
| :---: | :---: | :---: | :---: |
| I | 100 | 900 | 1 |
| II | 100 | 450 | 2 |
| III | 102 | 300 | 3 |
| IV | 102 | 150 | 6 |

Therefore, $S^{(1)}=100 \cdot 900+\frac{900 \cdot 899}{2} \cdot 1=$ $=90000+404550=494550$,

$S^{(2)}=100 \cdot 450+\frac{450 \cdot 449}{2} \cdot 2=247050$,

$$
\begin{aligned}
S^{(3)} & =102 \cdot 300+\frac{300 \cdot 299}{2} \cdot 3= \\
& =30600+134550=165150 \\
S^{(6)} & =102 \cdot 150+\frac{150 \cdot 149}{2} \cdot 6=82350 \\
S & =S^{(1)}-S^{(2)}-S^{(3)}+S^{(6)}=164700
\end{aligned}
$$
36. Find the sum of all positive three-digit numbers that are not divisible by 2 or 3.