For $n=2$, no matter how Nicuşor chooses $a_1$ and $a_2$, on the third row we will have two equal values. For $n=3$, suppose without loss of generality that $a_3=3$. Then, in the third column, we have only values equal to $3$. Since the set $(1, 2)$ admits only two permutations, among the first three rows there will be two identical ones, so these rows coincide and we cannot have any $a_i = 1$.

Next, we show that for $n \ge 4$, there exists a completion that works. If $n=4$ and we take $a_1 = 1, a_2 = 2, a_3 = 3, a_4 = 4$, the following construction satisfies the

Construction 1 (induction):
We will prove by induction on $n \ge 4$ that there exists a valid construction in which $a_i = i$ for every $i \in \{1, 2, \dots, n\}$ and the last row is exactly $1, 2, \dots, n$. The case $n = 4$ is illustrated above. Suppose there exists a valid table $T_n$ of size $(n+1) \times n$ for $n \ge 4$, for which the chosen permutation by Nicuşor is $a_1, a_2, \dots, a_n$ with $a_i = i$ for every $i = 1, 2, \dots, n$.
We construct a table $T_{n+1}$ of size $(n+2) \times (n+1)$, valid for $n+1$, as follows: we add a column at the end of $T_n$ and a new row between the last and the penultimate rows of $T_n$. We fill all the cells of the newly added last column with the value $n+1$, and the first $n$ values of the newly added row with $2, 3, \dots, n, 1$ (or any permutation that preserves the values for $a_1, a_2, \dots, a_n$).
It is easy to verify that the newly constructed table is valid.

Construction 2 (direct):
For $n \ge 4$, consider $a_i = i$ for every $i = 1, 2, 3, 4, \dots, n$ and the following construction:

It is easy to verify that the newly constructed table is valid.