Solution:

Answer: $212$

Consider the diagonals of the board running up and to the right - so the first diagonal is the square $1$, the second diagonal is the squares $2$ and $3$, and so on. The $i$th ascent is the largest step taken from a square in the $i$th diagonal to a square in the $i+1$st. Since you must climb from square $1$ to square $49$, the sum of the ascents is at least $48$. Since there are $12$ ascents, the average ascent is at least $4$.

The $1$st and $12$th ascents are at most $2$, and the $2$nd and $11$th ascents are at most $3$. The $6$th and $7$th ascents are at least $6$, and the $5$th and $8$th ascents are at least $5$. Because $f(x) = x^2$ is convex, the sum of squares of the ascents is minimized when they are as close together as possible. One possible shortest path is then $1 \rightarrow 3 \rightarrow 6 \rightarrow 10 \rightarrow 14 \rightarrow 19 \rightarrow 25 \rightarrow 31 \rightarrow 36 \rightarrow 40 \rightarrow 44 \rightarrow 47 \rightarrow 49$, which has ascents of size $2, 3, 4, 4, 5, 6, 6, 5, 4, 4, 3$, and $2$.

Thus, our answer is $212$, the sum of the squares of these ascents. There are other solutions to this problem. One alternative problem involves computing the shortest path to each square of the graph, recursively, starting from squares $2$ and $3$.