I have a ridiculous observation here: First of all consider the law of cosines for one of the angles in $\triangle ABC$: $a^2=c^2+b^2-2bc*cosA$ Since $a^2,b^2,c^2$ are all integers, we must have $2bc*cosA$(*) as an integer. Now by the area formula: $bc*sinA/2=1/2$ $\Rightarrow$ $bc=1/sinA$ We plug that into (*): (*)$=2\frac {cosA}{sinA}=2\sqrt {\frac {1-sin(A)^2}{sin(A)^2}}=2\sqrt {(\frac {1}{sinA})^2-1}$, which is an integer. So that means $(\frac {1}{sinA})^2-1$ must be a perfect square whole number. Since $(\frac {1}{sinA})^2$ is also perfect square ($\frac {1}{sinA}$ is forced to be an integer). Thus we must have $\frac {1}{sinA}=1$ ($1$ and $0$ are the only perfect squares that are separated by 1) or $A=90$. Analogously we can show that $B=C=90$, which is a clearly absurd result. Never mind I',m on the right track. $\frac {1}{sinA}$ doesn't have to be an integer!

Let $x \le y \le z$ be positive integers so that $\sqrt{x}, \sqrt{y}, \sqrt{z}$ are the side lengths of the triangle. By Heron's Formula, it's easy to obtain that $$2xy + 2yz + 2zx - x^2-y^2-z^2 = 4. \qquad (1)$$It's easy to check that $x+y \ge z$ implies that $x = y = 1, z = 2$. In view of this, we will prove that the desired placement of the triangle in the Cartesian plane exists by induction on $x+y+z.$ For the base case, if $x+y+z = 1 + 1 + 2 = 4$, then the placement is obvious. Otherwise, suppose that $x+y+z > 4.$ We know then that $z > x+y.$ Now, note that replacing $z$ with $2x+2y-z$ in $(1)$ still preserves the value of the LHS. Furthermore, as $2x+2y - z < z$, we can apply the inductive hypothesis on a triangle with side lengths $\sqrt{x}, \sqrt{y}, \sqrt{2x+2y-z}.$ Hence, let us consider $\triangle ABC$ with vertices at lattice points and $AB = \sqrt{x}, BC = \sqrt{y}, CA = \sqrt{2x+2y-z}.$ Now, consider the point $D$ such that $ABCD$ is a parallelogram. It is now easily verified that $\triangle BCD$ is the desired triangle, since $BD^2 = 2(AB^2 + BC^2) - AC^2 = z.$ The induction is complete.

Let $\triangle ABC$ be a triangle in the plane which is congruent to the paper triangle, which has three lattice points as vertices. This triangle exists by Step $1.$ Color the board with a checkerboard pattern, with each square of side length $\frac 12$, such that the origin is the lower-left corner of some square. We will say that two points in the Cartesian plane are equivalent if one can be sent to the other via reflections over lines of the form $x = a, y =b$ for integers $a, b \in \mathbb{Z}.$ For a point $P$ in the plane, let $\chi, \gamma$ be functions which map $P$ to its $x-$coordinate, $y-$coordinate respectively. Let $\alpha: \mathbb{R} \rightarrow \mathbb{R}$ be the function which maps a real number $r$ to the distance from $r$ to the nearest integer. It's easily verified that $P_1, P_2$ are equivalent if and only if $\alpha (\chi (P_1)) = \alpha (\chi(P_2))$ and $\alpha (\gamma(P_1)) = \alpha(\gamma(P_2)).$ We will require the following preliminary lemma. Lemma. For any two points $X, Y$ in the strict interior of $\triangle ABC,$ one of the points $A + \overrightarrow{XY}, A + \overrightarrow{YX}, B + \overrightarrow{XY}, B+ \overrightarrow{YX}, C+ \overrightarrow{XY}, C + \overrightarrow{YX}$ is either inside of on the perimeter of $\triangle ABC.$ Proof. If line $XY$ passes through one of the vertices, then the lemma is obvious. Otherwise, suppose WLOG that line $XY$ intersects segments $AB, AC$ and the extension of $BC$ past $C.$ Then it's easy to check that one of $C + \overrightarrow{XY}, C + \overrightarrow{YX}$ is on the perimeter of or inside $\triangle ABC.$ $\blacksquare$ Claim. No two interior points of $\triangle ABC$ are of the same color and equivalent, where here "interior" does not include the edges of the triangle. Proof. We will call a point a mattice point if it is the midpoint of a segment connecting two lattice points. It is then easily checked by Shoelace, say, that $\triangle ABC$ has no mattice points in its strict interior. Furthermore, the only mattice points on its perimeter are $A, B, C$ and the midpoints of $AB, BC, CA.$ Finally, it's also clear that the only lattice points inside or on the perimeter of $\triangle ABC$ are $A, B, C.$ Consider two arbitrary points $P_1, P_2$ in the plane which are the same color and equivalent. Suppose, for contradiction, that both lied in the strict interior of $\triangle ABC$. By the previous observation, we may assume that $P_1, P_2$ are not mattice points. We consider two cases. Case 1. $\chi (P_1) + \chi (P_2), \gamma(P_1) + \gamma(P_2) \in \mathbb{Z}$ Then, the midpoint of $P_1P_2$ is a mattice point. Furthermore, this mattice point must lie strictly within $\triangle ABC$. However, this is clearly a contradiction. Case 2. One of $\chi(P_1) + \chi(P_2), \gamma(P_1) + \gamma(P_2)$ is not integral It's the easy to see with the same color condition that $\chi(P_1) - \chi(P_2), \gamma(P_1) - \gamma(P_2)$ are both integers. In other words, $\overrightarrow{P_1P_2}$ has both components integral. By the Lemma, we can assume by symmetry that $A + \overrightarrow{P_1P_2}$ is inside or on the perimeter of $\triangle ABC.$ However, as this is a lattice point, we know that this must be $B$ or $C.$ This is clearly impossible if $P_1, P_2$ are in the strict interior of $\triangle ABC,$ contradiction. As we've exhausted all cases, the claim is proven. $\blacksquare$ We will now attempt to describe the algorithm. Start by selecting placing the cardboard square on an arbitrary white square in the plane so that it is on top of some part of $\triangle ABC.$ At some point in time, define a move on the triangle as follows. First, select a side of the cardboard square, say the top side for example. Then, reflect the part of the triangle which is above the top side over the top side. If this part was previous below the cardboard square, then this move will bring it above the cardboard square. Otherwise, this move would bring it below the cardboard square. Notice that these moves preserve the property that no point in the strict interior of the triangle will ever be on top of a mattice point. Furthermore, notice that the only way for a point of the triangle to be directly on top of the square at any point in time is if it began on a black square, and the only way for a point of the triangle to be directly on top of the square at any point in time is if it began on a white square. Also, observe that no black point in the strict interior of the triangle can ever be directly on top of another black point in the strict interior of the triangle, as this would imply that they were equivalent and therefore violate the claim. To finish, it suffices only to observe that the area of the triangle which is directly on top of or directly below the cardboard square is strictly increasing with each move, so if we repeated apply moves we will reach a state where every point of the triangle is either directly above or directly below the cardboard square. Finally, use the previous observations to conclude that no point on the square is covered twice, and so the fact that the surface area of the cardboard square is the same as the area of the triangle implies that we've obtained our desired folding.