The least $n$ in question is $1022$. An example that $n = 1021$ is not enough: arrange in a row $32$ blocks $1, 1, \ldots, 1, 32$, with $30$ ones in each, then add $29$ ones. This gives a sequence of length $32 \cdot 31 + 29 = 1021$ and sum $62 \cdot 32 + 29 = 2013$. No consecutive terms in it have sum $31$.

Take a sequence with positive integer terms with length $1022$ and sum $2013$. We show that several consecutive terms add up to $31$. Let $S_j$ be the sum of its first $j$ terms, $j = 1, \ldots, 1022$. Consider the two sequences
$$1 \leq S_1 < S_2 < \dots < S_{1022} = 2013 \text{ and } 32 \leq 31 + S_1 < 31 + S_2 < \dots < 31 + S_{1022} = 2044.$$
Assume $S_j \neq 31$ for all $j$, otherwise the claim follows. Since $31 + S_j \neq 31$ for all $j$, it follows that the $2044$ integers above are in $[1, 2044]$ and different from $31$. Hence two of them are the same. Because the terms in each sequence are distinct, there are indices $i, j$ such that $S_i = S_j + 31$. Clearly $i > j$ and the terms of the original sequence with indices $j+1, \ldots, i$ have sum $31$.