Integer Linear Programming Without using a computer, I have to solve the following integer linear programming:$$\min \quad x_1+x_2+x_3$$
$$\operatorname{sub} :\begin{cases}x_1\le9\\x_2\le7\\x_3\le5\\3x_1+6x_2+8x_3=80\\x_1,x_2,x_3\in\mathbb{N}\end{cases}$$
Is there any algebraic method to compute the solution (I can't use the simplex method)? 

Yep, Peter Shor got the right approach. You also need to use divisibility concepts to reduce the amount of work. Start with $x_3=5$. This would mean that $3x_1+6x_2=40$. But this can't lead to a solution because the left-hand side is divisible by 3, but the right-hand side is not! So try $x_3=4$, which gives $3x_1+6x_2=48$, which is divisible by $3$. Now try the maximum value for $x_2$, which is 7 and that gives $x_1= 2$.
This may or may not be convincing enough as proof, depending how formally the greedy approach needs to justified. The intuition is that $x_3$ contributes the most to the equality constraint, $x_1$ the least, while their appearance in the objective function is equal, so one unit of $x_3$ "buys" 8/3 units of $x_1$ in the objective function.
But clearly, the only other choice for $x_3$ (divisibility-wise) is 1, and that can't be met by plugging even the maximum values in $x_1+2x_2=25$; the max you can get maxing both $x_1=5$ and $x_2=7$ is 19.
A similar greedy argument goes for maxing $x_2$ instead of $x_1$ (once $x_3 = 4$ is settled); One unit of $x_2$ buys two units of $x_1$ in the objective function. If instead of $x_1 = 2$, $x_2=7$ you satisfy $x_1+2x_2=16$ with $x_1=4, x_2=6$, the objective function increases (by 1).
Alternatively, this remainder optimization problem being in only two variables ($x_1$, $x_2$), you can solve it by graphing its polyhedron. 


The fact that the greedy method can solve certain classes of I[L]Ps can be given quite a formal proof(s) of correctenss; see the paper by V.V. Shenmaier for example, although I don't dream to suggest that's going to be accessible to an 11-year old.
