Let the numbers in a row be $a_1, a_2, \dots, a_{125}$. By conditions, we have $a_{i+1} > \frac{a_i + a_{i+2}}{2}$ for every $i = 1, 2, \dots, 123$, which is equivalent to $a_{i+1} - a_i > a_{i+2} - a_{i+1}$. Denoting $d_i = a_{i+1} - a_i$, we have $d_1 > d_2 > \dots > d_{124}$.

Let $a_m$ be the largest among $a_1, \dots, a_{125}$; then $d_1, d_2, \dots, d_{m-1}$ are positive and $d_m, d_{m+1}, \dots, d_{124}$ are negative. If both $1$ and $-1$ occurred among the differences $d_1, d_2, \dots, d_{124}$, then they should be consecutive, i.e., $d_i = 1$ and $d_{i+1} = -1$ for some $i$, whence $a_{i+1} = a_{i-1}$. Contradiction to the assumption that the given numbers are distinct shows that either $1$ or $-1$ is missing among the differences. W.l.o.g., assume that $1$ is missing (if $-1$ is missing, we can reverse the numeration of the given numbers).

Then $a_m = a_1 + (d_1 + d_2 + \dots + d_{m-1}) \ge 1 + (m + (m-1) + \dots + 2) = 1 + 2 + \dots + m$ and $a_m = a_{125} - (d_m + d_{m+1} + \dots + d_{124}) \ge 1 + (1 + 2 + \dots + (125 - m))$. Among numbers $m$ and $125 - m$, one is at least $63$, whence the previously established inequalities imply $a_m \ge 1 + 2 + \dots + 63$.

On the other hand, taking $a_1 = 1$ and $d_i = 64 - i$, $1 \le i \le 62$, and $d_i = 62 - i$, $63 \le i \le 124$, the largest number $a_{63}$ equals $1 + 2 + \dots + 63$, while all numbers $a_i$ are positive, as $1 = a_1 < a_2 < \dots < a_{63}$ and $a_{63} > a_{64} > \dots > a_{125} = (1 + 2 + \dots + 63) - (1 + 2 + \dots + 62) = 63$. All these numbers are distinct, since for every $i = 64, 65, \dots, 125$, $a_i = (1 + 2 + \dots + 63) - (1 + 2 + \dots + (i - 63)) = (i - 62) + (i - 61) + \dots + 63 = a_{127 - i} - 1$, whence $a_i$ lies strictly between $a_{126 - i}$ and $a_{127 - i}$. Consequently, the largest number written in the row is $1 + 2 + \dots + 63$, i.e., $2016$.