For any finite-energy signal $x(t)$, the function
$$y(t) = \sum_{n=-\infty}^\infty x(t-nT), -\infty < t < \infty$$
is a periodic function of $t$ with period $T$, that is, $y(t-T) = y(t), -\infty < t < \infty$. Note that $T$ is not necessarily the least period of $y(t)$. Now, a periodic
function can be expressed as a Fourier series
$$y(t) = \sum_{m=-\infty}^\infty c_m \exp(j2\pi mt/T)$$ where
$$\begin{align}
c_m &= \frac{1}{T}\int_0^T y(t)\exp(-j2\pi mt/T)\,\mathrm dt\\
&= \frac{1}{T}\int_0^T \sum_{n=-\infty}^\infty x(t-nT)\exp(-j2\pi mt/T)\,\mathrm dt\\
&= \frac{1}{T}\sum_{n=-\infty}^\infty \int_0^T x(t-nT)\exp(-j2\pi mt/T)\,\mathrm dt\\
&= \frac{1}{T}\sum_{n=-\infty}^\infty \int_{-nT}^{-(n-1)T} x(s)\exp(-j2\pi ms/T)\,\mathrm ds\\
&= \frac{1}{T}\int_{-\infty}^{\infty} x(s)\exp(-j2\pi ms/T)\,\mathrm ds\\
&= \frac{1}{T}X\left(\frac{m}{T}\right)
\end{align}$$
where $X(f)$ is the Fourier transform of $x(t)$.
Now, for the case $T = 1$,
if $y(t) = 1$ for all $t$ as the OP desires, then it must be that $c_0 = 1$
and $c_m = 0$ for all $m\neq 0$, and therefore the Fourier transform of $x(t)$
must have the property that
$$X(m) = \begin{cases} 1, & m = 0,\\ 0, &m \neq 0,\end{cases}\tag{1}$$
that is, $X(f)$ is what is sometimes referred to as a Nyquist pulse (in the
frequency domain). We are familiar with Nyquist pulses in the time domain,
with $\operatorname{sinc}(t) = \frac{\sin(\pi t)}{\pi t}$ being the most
familiar one, but there are many others, including as a special case
$[\operatorname{sinc}(t)]^k$. Applying duality,
$X(f) = [\operatorname{sinc}(f)]^k$ is a Nyquist
pulse in the frequency domain, and its inverse Fourier transform $x(t)$ is
a signal with the property that
$$\sum_{n=-\infty}^\infty x(t-nT) = 1, -\infty < t < \infty.$$
Note that $x(t)$ is the $k$-fold convolution of $\operatorname{rect}(t)$
with itself. For example, with $k = 2$,
$$x(t) = \begin{cases}1 - |t|, &-1 < t < 1,\\
0, & \text{otherwise.}\end{cases}$$
What about the other way around? Well, $\operatorname{sinc}(t)$ has
Fourier transform $\operatorname{rect}(f)$ which has the Nyquist pulse
property $(1)$ in the frequency domain as does $[\operatorname{sinc}(t)]^2$
whose Fourier transform
$$X(f) = \begin{cases}1 - |f|, &-1 < f < 1,\\
0, & \text{otherwise,}\end{cases}$$
also satisfies $(1)$. However, for $k > 2$,
the Fourier transform of $[\operatorname{sinc}(t)]^k$ does not
have the Nyquist pulse property $(1)$ because the three-fold (or more
generally $k$-fold convolution of $\operatorname{rect}(f)$ with itself
does not satisfy $(1)$: the $k$-fold convolution "bulges" outside
the interval $(-1,1)$. See for, example, the
Irwin-Hall distribution
for details of the exact values of the convolution results.
