It suffices to show that there exists a Hamiltonian cycle with edge sum equal to $\pm 1$, as we can remove the appropriate edge from the cycle to make the edge sum of the path zero. We now show that this cycle exists. Clearly a cycle with odd length has odd sum. First we partition the edges of the graph into 50 Hamiltonian cycles. Claim: We can partition the graph into 50 Hamiltonian cycles. Proof: Enumerate the vertices 1 to 101. Then for $1 \leq k \leq 50$, consider the cycle with edges $(i,i+k)$. Since 101 is prime, this always results in a cycle of 101 edges. $\square$ Now we show that some $C_{101}$ in the graph has sum $\pm 1$. Suppose otherwise, then the cycles we have constructed are all at least $3$, or at most $-3$. Note that since there are 50 cycles, it cannot be the case that all of them are greater than $3$, or all less than $-3$ (due to the restriction that the absolute value of the sum of numbers on all the edges is less than $150$). Thus there exists a cycle $C_1$ with sum less than $3$, and a cycle $C_2$ with sum greater than $-3$. The idea is to transform $C_1$ to $C_2$ in a sequence of moves so that one of the intermediate cycles has sum equal to $\pm 1$. Now the cycle $C_2$ can be obtained from $C_1$ by a sequence of operations, where we swap two consecutive vertices of a cycle. Each operation changes the sum total of the edges in the cycle by at most 4, as two edges are replaced by two other edges. Thus at some point there is an intermediate cycle with sum total $\pm 1$.