7. The line $l: x+y=t$ intersects the circle $\odot O: x^{2}+y^{2}=20$ at points $A$ and $B$, and $S_{\triangle O A B}$ is an integer. Then the number of all positive integer values of $t$ that satisfy the condition is $\qquad$ .
7.2.

Let $\angle A O B=2 \alpha, O C \perp A B$, with the foot of the perpendicular being $C$. Then $O C=\sqrt{20} \cos \alpha=\frac{t}{\sqrt{1+1}}=\frac{t}{\sqrt{2}}$ $\Rightarrow \cos \alpha=\frac{t}{2 \sqrt{10}}$.
Also, $S_{\triangle O A B}=10 \sin 2 \alpha \leqslant 10$, so $\sin 2 \alpha=\frac{k}{10}(k \in\{1,2, \cdots, 10\})$.
Thus, $2 \cos \alpha \sqrt{1-\cos ^{2} \alpha}=\frac{k}{10}$.
Substituting equation (1) into the above equation and simplifying, we get $t^{2}\left(40-t^{2}\right)=4 k^{2}$.
It is easy to see that $t$ is an even number. Let $t=2 t_{0}$. Then $4 t_{0}^{2}\left(10-t_{0}^{2}\right)=k^{2} \Rightarrow t_{0}<\sqrt{10}$. Upon inspection, $t_{0}=1,3$ satisfy the conditions. Therefore, $t=2,6$.