Let $N$ be a convex polygon with 1415 vertices and perimeter 2001. Prove that we can find 3 vertices of $N$ which form a triangle of area smaller than 1.
1. **Determine the longest side:**
   Given that the perimeter of the convex polygon \( N \) with 1415 vertices is 2001, the longest side must have a length of at least:
   \[
   \frac{2001}{1415}
   \]
   This is because if all sides were equal, each side would be \(\frac{2001}{1415}\). Since the sides are not necessarily equal, the longest side must be at least this length.

2. **Calculate the total length of the remaining sides:**
   If the longest side is at least \(\frac{2001}{1415}\), the total length of the remaining 1414 sides is at most:
   \[
   2001 - \frac{2001}{1415} = 2001 \left(1 - \frac{1}{1415}\right) = 2001 \times \frac{1414}{1415}
   \]

3. **Pair the remaining sides:**
   Divide the 1414 remaining sides into 707 pairs of consecutive sides. The sum of the lengths of each pair of consecutive sides is at most:
   \[
   \frac{2001 \times 1414}{1415 \times 707} = \frac{2001 \times 2}{1415} = \frac{4002}{1415} < 2\sqrt{2}
   \]
   This inequality holds because:
   \[
   \frac{4002}{1415} \approx 2.828 < 2\sqrt{2} \approx 2.828
   \]

4. **Consider a pair of consecutive sides \(AB\) and \(BC\):**
   There exist two consecutive sides, say \(AB\) and \(BC\), such that:
   \[
   AB + BC \leq \frac{4002}{1415} < 2\sqrt{2}
   \]

5. **Calculate the area of triangle \(ABC\):**
   Let \(S\) be the area of triangle \(ABC\). Using the formula for the area of a triangle with sides \(a\) and \(b\) and included angle \(\theta\):
   \[
   S = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(\theta)
   \]
   Using the AM-GM inequality, we have:
   \[
   \sqrt{S} = \sqrt{\frac{1}{2} \cdot AB \cdot BC \cdot \sin(\theta)} \leq \frac{1}{\sqrt{2}} \sqrt{AB \cdot BC}
   \]
   Since \(AB + BC < 2\sqrt{2}\), we have:
   \[
   \sqrt{AB \cdot BC} \leq \frac{AB + BC}{2} < \frac{2\sqrt{2}}{2} = \sqrt{2}
   \]
   Therefore:
   \[
   \sqrt{S} \leq \frac{1}{\sqrt{2}} \cdot \sqrt{2} = 1
   \]
   Squaring both sides, we get:
   \[
   S < 1
   \]

Conclusion:
\[
\boxed{S < 1}
\]