Determine all points on the straight line which joins $(4, 11)$ to $(16, 1)$ and whose coordinates are positive integers.

1. **Find the slope of the line:**
   The slope \( m \) of the line passing through the points \((4, 11)\) and \((16, 1)\) is given by:
   \[
   m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 11}{16 - 4} = \frac{-10}{12} = -\frac{5}{6}
   \]

2. **Find the equation of the line:**
   Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), and substituting \( m = -\frac{5}{6} \) and the point \((4, 11)\):
   \[
   y - 11 = -\frac{5}{6}(x - 4)
   \]
   Simplifying this equation:
   \[
   y - 11 = -\frac{5}{6}x + \frac{20}{6}
   \]
   \[
   y - 11 = -\frac{5}{6}x + \frac{10}{3}
   \]
   \[
   y = -\frac{5}{6}x + \frac{10}{3} + 11
   \]
   \[
   y = -\frac{5}{6}x + \frac{10}{3} + \frac{33}{3}
   \]
   \[
   y = -\frac{5}{6}x + \frac{43}{3}
   \]

3. **Convert to standard form:**
   To find integer solutions, convert the equation to standard form:
   \[
   6y = -5x + 43
   \]
   \[
   5x + 6y = 43
   \]

4. **Find integer solutions:**
   We need to find positive integer solutions \((x, y)\) such that \( 5x + 6y = 43 \). We can solve for \( y \) in terms of \( x \):
   \[
   6y = 43 - 5x
   \]
   \[
   y = \frac{43 - 5x}{6}
   \]
   For \( y \) to be an integer, \( 43 - 5x \) must be divisible by 6. We check values of \( x \) that make \( 43 - 5x \) a multiple of 6:
   \[
   43 - 5x \equiv 0 \pmod{6}
   \]
   \[
   43 \equiv 5x \pmod{6}
   \]
   \[
   1 \equiv 5x \pmod{6}
   \]
   The multiplicative inverse of 5 modulo 6 is 5 (since \( 5 \times 5 \equiv 1 \pmod{6} \)), so:
   \[
   x \equiv 5 \pmod{6}
   \]
   Therefore, \( x = 6k + 5 \) for integer \( k \). We need \( x \) to be a positive integer:
   \[
   x = 5, 11, 17, \ldots
   \]
   We substitute these values back into the equation to find corresponding \( y \) values:
   - For \( x = 5 \):
     \[
     y = \frac{43 - 5 \times 5}{6} = \frac{43 - 25}{6} = \frac{18}{6} = 3
     \]
   - For \( x = 11 \):
     \[
     y = \frac{43 - 5 \times 11}{6} = \frac{43 - 55}{6} = \frac{-12}{6} = -2 \quad (\text{not a positive integer})
     \]

   Thus, the only positive integer solution is \((5, 3)\).

The final answer is \(\boxed{(5, 3)}\).