Solution

The author suggests two ways to produce a proton-antiproton pair. We interpret the conditions as in the second case the electron and positron move toward each other with equal speeds.

We also remember that the minimum energy corresponds to the final particles being at rest relative to each other. Then the invariant in the first case is:

$$I_1 = \frac{(E_{e} + m_e c^2)^2}{c^2} - p_e^2 = \frac{(2m_p c^2)^2}{c^2} = 4m_p^2 c^2 \tag{1}$$

where $E_{e} = m_e c^2+E_{k1}$ and $p_e = \gamma m_e v = \frac{\beta}{c} E_e$. One could set up complicated equations in terms of $\beta$, but it is easier to write another invariant, this time for the positron:

$$m_e^2 c^2 = \frac{E_e^2}{c^2} - p_e^2 \tag{2}$$

Substituting (2) into (1):

$$\frac{(E_e + m_e c^2)^2}{c^2} - \frac{E_e^2}{c^2} = 4m_p^2 c^2 - m_e^2 c^2$$

$$2E_e m_e c^2 = 4m_p^2 c^4 - 2m_e^2 c^4$$

$$E_e = c^2 \frac{2m_p^2 - m_e^2}{m_e}$$

$$E_{k1} = E_e - m_e c^2 = 2c^2 \frac{m_p^2 - m_e^2}{m_e} \tag{3}$$

Now we consider the second case similarly, noting that the total momentum of the system is zero:

$$I_2 = \frac{(2E_e)^2}{c^2} = \frac{(2m_p c^2)^2}{c^2} = 4m_p^2 c^2 \tag{4}$$

$$E_e = m_p c^2 \tag{5}$$

$$E_{k2} = c^2 (m_p - m_e)$$

$$N = \frac{E_{k1}}{E_{k2}} = 2 \frac{m_p^2 - m_e^2}{m_e (m_p - m_e)} = 2 \left( \frac{m_p}{m_e} + 1 \right) \approx 3.7 \cdot 10^3$$
It is clear that almost all the kinetic energy in the first case goes not to the creation of the pair, but to its acceleration.

Note: Savchenko has a typo in the second part of the answer, but it hardly affects the numerical result.