Problem 1326

1. Let's denote $x = 1995$. We need to compare the two fractions:
$$\frac{x^{1994} + 1}{x^{1995} + 1} \quad \text{and} \quad \frac{x^{1995} + 1}{x^{1996} + 1}$$

2. Consider the function $f(x) = \frac{x^n + 1}{x^{n+1} + 1}$. We need to show that this function is decreasing for $x > 0$.

3. To show that $f(x)$ is decreasing, we can compute its derivative and show that it is negative for $x > 0$.

4. The derivative of $f(x)$ with respect to $x$ is:
$$f'(x) = \frac{d}{dx} \left( \frac{x^n + 1}{x^{n+1} + 1} \right)$$

5. Using the quotient rule, we have:
$$f'(x) = \frac{(x^{n+1} + 1) \cdot \frac{d}{dx}(x^n + 1) - (x^n + 1) \cdot \frac{d}{dx}(x^{n+1} + 1)}{(x^{n+1} + 1)^2}$$

6. Simplifying the derivatives:
$$\frac{d}{dx}(x^n + 1) = n x^{n-1}$$
$$\frac{d}{dx}(x^{n+1} + 1) = (n+1) x^n$$

7. Substituting these into the derivative expression:
$$f'(x) = \frac{(x^{n+1} + 1) \cdot n x^{n-1} - (x^n + 1) \cdot (n+1) x^n}{(x^{n+1} + 1)^2}$$

8. Simplifying the numerator:
$$f'(x) = \frac{n x^{n-1} x^{n+1} + n x^{n-1} - (n+1) x^n x^n - (n+1) x^n}{(x^{n+1} + 1)^2}$$
$$f'(x) = \frac{n x^{2n} + n x^{n-1} - (n+1) x^{2n} - (n+1) x^n}{(x^{n+1} + 1)^2}$$
$$f'(x) = \frac{n x^{2n} + n x^{n-1} - n x^{2n} - x^{2n} - (n+1) x^n}{(x^{n+1} + 1)^2}$$
$$f'(x) = \frac{- x^{2n} + n x^{n-1} - (n+1) x^n}{(x^{n+1} + 1)^2}$$

9. Since $x > 0$, the term $- x^{2n}$ dominates, making $f'(x) < 0$. Therefore, $f(x)$ is a decreasing function for $x > 0$.

10. Applying this result to our specific case with $x = 1995$, $n = 1994$, and $n+1 = 1995$:
$$\frac{1995^{1994} + 1}{1995^{1995} + 1} > \frac{1995^{1995} + 1}{1995^{1996} + 1}$$

Conclusion:
$$\boxed{\frac{1995^{1994} + 1}{1995^{1995} + 1} > \frac{1995^{1995} + 1}{1995^{1996} + 1}}$$