First, it is a sufficient condition.
As long as $f$ is bounded on $[0,1]$, the upper and lower sums corresponding to arbitrary partitions are bounded. Let $\mathcal{P}$ denote the set of all partitions of $[0,1].$ Consequently, the sets $\{L(P,f): P \in \mathcal{P}\}$ and $\{U(P,f): P \in \mathcal{P}\}$ are bounded, and this guarantees the existence of
$$\underline{\int}_0^1 f(x) \, dx = \sup_{P \in \mathcal{P}}\, L(P,f), \\ \overline{\int}_0^1 f(x) \, dx = \inf_{P \in \mathcal{P}}\, U(P,f) , $$
which are called the lower and upper integrals.
Given any regular partition $D_n$ we have
$$L(D_n,f) \leqslant \underline{\int}_0^1 f(x) \, dx \leqslant \overline{\int}_0^1 f(x) \, dx \leqslant U(D_n,f).$$
The central inequality follows because for any partitions $P$ and $Q$ we have $L(P,f) \leqslant U(Q,f)$ (take a common refinement of the partitions to show this) and, thus $\sup_{P \in \mathcal{P}} \,L(P,f) \leqslant \inf_{Q \in \mathcal{P}} \,U(Q,f)$.
Hence,
$$0 \leqslant \overline{\int}_0^1 f(x) \, dx - \underline{\int}_0^1 f(x) \, dx \leqslant U(D_n,f) - L(D_n,f).$$
The right-hand side converges to $0$ as $n \to \infty$, by hypothesis, which implies that $f$ is integrable since we must have
$$\underline{\int}_0^1 f(x) \, dx = \overline{\int}_0^1 f(x) \, dx, $$
where the common value of lower and upper integrals is by definition the value of the integral.
To show it is a necessary condition, consider
$$\left|U(D_n,f) - L(D_n,f) \right| \leqslant \left|U(D_n,f) - \int_0^1 f(x) \, dx \right| + \left|L(D_n,f) - \int_0^1 f(x) \, dx \right|.$$
The two terms on the RHS go to zero as $n \to \infty$. This is a consequence of the equivalent condition for integrability where for arbitrary Riemann sums corresponding to tagged partitions we have
$$\tag{*}\int_0^1 f(x) \, dx = \lim_{\|P\| \to 0} S(P,f).$$
Here $\|P\| = \max_{1 \leqslant j \leqslant n} (x_j - x_{j-1})$ is the norm of the partition $P = (x_0,x_1, \ldots, x_n)$ and, clearly, $\|D_n\| \to 0$ if and only if $n \to \infty$.
It takes a bit of effort to prove the equivalence of $(*)$ to the definition of the Riemann integral in terms of partition refinement or the Darboux approach. It has been shown a number of times on this site including here.
Best Answer
Yes, each lower sum is $0$. And the upper sum that corresponds to your partition is $\frac{18}n$. Since $\lim_{n\to\infty}\frac{18}n=0$…