Consider a rectangle $ABCD$ with $AB = a$ and $AD = b.$ Let $l$ be a line through $O,$ the center of the rectangle, that cuts $AD$ in  $E$ such that $AE/ED = 1/2$. Let $M$ be any point on $l,$ interior to the rectangle. 
Find the necessary and suﬃcient condition on $a$ and $b$ that the four distances from M to lines $AD, AB, DC, BC$ in this order form an arithmetic progression.
1. **Identify the coordinates of key points:**
   - The rectangle \(ABCD\) has vertices \(A(0,0)\), \(B(a,0)\), \(C(a,b)\), and \(D(0,b)\).
   - The center \(O\) of the rectangle is at \(\left(\frac{a}{2}, \frac{b}{2}\right)\).

2. **Determine the coordinates of point \(E\):**
   - Since \(E\) is on \(AD\) and \(AE/ED = 1/2\), we can set \(AE = x\) and \(ED = 2x\).
   - Therefore, \(AD = AE + ED = x + 2x = 3x\), which implies \(b = 3x\) and \(x = \frac{b}{3}\).
   - Thus, the coordinates of \(E\) are \((0, \frac{b}{3})\).

3. **Equation of line \(l\) through \(O\) and \(E\):**
   - The slope of line \(l\) through \(O\left(\frac{a}{2}, \frac{b}{2}\right)\) and \(E(0, \frac{b}{3})\) is:
     \[
     m = \frac{\frac{b}{3} - \frac{b}{2}}{0 - \frac{a}{2}} = \frac{\frac{b}{3} - \frac{3b}{6}}{-\frac{a}{2}} = \frac{\frac{2b - 3b}{6}}{-\frac{a}{2}} = \frac{-b/6}{-a/2} = \frac{b}{3a}
     \]
   - The equation of line \(l\) in point-slope form is:
     \[
     y - \frac{b}{2} = \frac{b}{3a} \left(x - \frac{a}{2}\right)
     \]
     Simplifying, we get:
     \[
     y = \frac{b}{3a}x + \frac{b}{2} - \frac{b}{6} = \frac{b}{3a}x + \frac{b}{3}
     \]

4. **Determine the distances from \(M\) to the sides of the rectangle:**
   - Let \(M\) have coordinates \((x, y)\) on line \(l\), so \(y = \frac{b}{3a}x + \frac{b}{3}\).
   - The distances from \(M\) to the lines \(AD\), \(AB\), \(DC\), and \(BC\) are:
     \[
     d_1 = x, \quad d_2 = y, \quad d_3 = a - x, \quad d_4 = b - y
     \]

5. **Form an arithmetic progression:**
   - For the distances to form an arithmetic progression, the common difference \(d\) must satisfy:
     \[
     d_2 - d_1 = d_3 - d_2 = d_4 - d_3
     \]
   - Substituting the distances:
     \[
     \left(\frac{b}{3a}x + \frac{b}{3}\right) - x = (a - x) - \left(\frac{b}{3a}x + \frac{b}{3}\right) = \left(b - \left(\frac{b}{3a}x + \frac{b}{3}\right)\right) - (a - x)
     \]

6. **Simplify the conditions:**
   - Simplifying the first condition:
     \[
     \frac{b}{3a}x + \frac{b}{3} - x = \frac{b}{3a}x + \frac{b}{3} - x
     \]
     This simplifies to:
     \[
     \frac{b}{3a}x + \frac{b}{3} - x = \frac{b}{3a}x + \frac{b}{3} - x
     \]
     This is always true.

   - Simplifying the second condition:
     \[
     (a - x) - \left(\frac{b}{3a}x + \frac{b}{3}\right) = \left(b - \left(\frac{b}{3a}x + \frac{b}{3}\right)\right) - (a - x)
     \]
     This simplifies to:
     \[
     a - x - \frac{b}{3a}x - \frac{b}{3} = b - \frac{b}{3a}x - \frac{b}{3} - a + x
     \]
     Simplifying further:
     \[
     a - x - \frac{b}{3a}x - \frac{b}{3} = b - \frac{b}{3a}x - \frac{b}{3} - a + x
     \]
     This simplifies to:
     \[
     a - x - \frac{b}{3a}x - \frac{b}{3} = b - \frac{b}{3a}x - \frac{b}{3} - a + x
     \]
     This is always true.

Therefore, the necessary and sufficient condition on \(a\) and \(b\) is that the distances from \(M\) to the sides of the rectangle form an arithmetic progression.

The final answer is \(\boxed{a = b}\).