Number of samples in one period of subcarrier after IFFT
If you'd like to think analog, an OFDM signal can be written as a sum of weighted complex sinusoidals:
$$
x(t)=\sum_{k=0}^{N-1}X_k \exp{\left( \mathrm j 2\pi\frac{kf_\mathrm{s}}{N}t \tag{1}\right)},
$$
where $N$ is the number of subcarriers, $f_\mathrm{s}$ is the sampling frequency and $X_k$ denotes the subcarrier values. For a digital implementation, (1) is sampled at $t=n/f_\mathrm{s}$:
$$
x(n/f_\mathrm{s})=\sum_{k=0}^{N-1}X_k \exp{\left( \mathrm j 2\pi\frac{kn}{N} \tag{2}\right)},
$$
which corresponds to the inverse discrete Fourier transform (IDFT), implemented by an IFFT. The IFFT has $N$ input values and $N$ output values. That's ok, because the IDFT is N-periodic and thus any additional sample would be redundant. Consequently, every subcarrier in (1) is represented by $N$ samples, regardless of its number of periods during the OFDM symbol length.

If you try to imagine the digital OFDM time domain signal as a superposition of $N$ sampled sinusoidals, the figure you posted has two problems, in my opinion: 


The first subcarrier in the image obviously has frequency $f_\mathrm{s}/N$ and the fourth subcarrier has frequency $f_\mathrm{s}$. That's wrong. From (1) you can see that the first subcarrier should have frequency 0 and the fourth should have frequency $(N-1)f_\mathrm{s}/N$. You can reorder the subcarriers (e.g. from negative to positive) but the DC subcarrier must be included and there must not exist a subcarrier with frequency $f_\mathrm{s}$. (One could argue that subcarriers with frequency 0 and $f_\mathrm{s}$ are actually the same but I think it's a poor choice for an illustration)
The image only shows the real (or imaginary) part of the sinusoidals. That could give you the impression that some aliasing is going on when sampling. I will explain in a second what I mean.


Please consider the following figure that I have created instead


It contains the subcarriers from $k=0$ to $k=3$ (lines) and their samples (dots). It also shows the real part (blue) and imaginary part (red). Now consider the red line in the first and the third subfigure: their samples are exactly the same. However, the blue (imaginary) part is different. That is, $N=4$ samples are sufficient here to represent all subcarriers.

Here is the Matlab code I used to create the above figure:

N = 4; % num subcarriers
os = 128; % oversampling
t = linspace(0, 1-1/(N*os), N*os); % time
n = 0:N-1; % discrete time

figure;
cnt = 1; % helper
for k = 0:N-1;
    subcarrier = exp(1i * 2*pi*k*t);

    subplot(N,1,cnt);
    plot( t, real(subcarrier), 'b-');
    hold on;
    plot( t(n*os+1), real(subcarrier(n*os+1)), 'b.', 'MarkerSize', 20);

    plot( t, imag(subcarrier), 'r--');
    hold on;
    plot( t(n*os+1), imag(subcarrier(n*os+1)), 'r.', 'MarkerSize', 20);

    if(k ~= N-1)
        set(gca, 'XTick', []);
    else
        xlabel('t/T');
    end
    ylim([-1 1]);
    title( sprintf('k=%d', k));

    cnt = cnt + 1;
end