Let $x$ be the length of one side of a triangle and let $y$ be the height to that side.  If $x+y=418$, find the maximum possible $\textit{integral value}$ of the area of the triangle.
1. We start with the given equation \( x + y = 418 \). We need to maximize the area of the triangle, which is given by \( \frac{1}{2}xy \).

2. To find the maximum value of \( \frac{1}{2}xy \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. According to the AM-GM inequality:
   \[
   \frac{x + y}{2} \geq \sqrt{xy}
   \]
   Substituting \( x + y = 418 \) into the inequality, we get:
   \[
   \frac{418}{2} \geq \sqrt{xy}
   \]
   Simplifying, we have:
   \[
   209 \geq \sqrt{xy}
   \]

3. Squaring both sides to eliminate the square root, we obtain:
   \[
   209^2 \geq xy
   \]
   Calculating \( 209^2 \):
   \[
   209^2 = 43681
   \]
   Therefore:
   \[
   43681 \geq xy
   \]

4. The area of the triangle is given by \( \frac{1}{2}xy \). To find the maximum possible area, we use the maximum value of \( xy \):
   \[
   \frac{1}{2}xy \leq \frac{1}{2} \times 43681
   \]
   Simplifying, we get:
   \[
   \frac{1}{2} \times 43681 = 21840.5
   \]

5. Since we are asked for the maximum possible integral value of the area, we take the floor of 21840.5:
   \[
   \lfloor 21840.5 \rfloor = 21840
   \]

Conclusion:
The maximum possible integral value of the area of the triangle is \( \boxed{21840} \).