Ambiguity and confusion can be avoided if one uses unit vector notation as an intermediate step and then remove them.
Assuming down as negative and up as positive
##\mathbf F_{\text{th.}}+\mathbf F_{\text{g}}=M\dfrac{\Delta \mathbf{v}}{\Delta t}.##
## F_{\text{th.}}~(\mathbf{\hat{y}})+(245,000~\text{N})~(-\mathbf{\hat{y}})=(25,000~\text{kg})\dfrac{0-(172~\text{m/s})(-\mathbf{\hat{y}}) }{(13~\text{s})}.##
## F_{\text{th.}}-(245,000~\text{N})=(25,000~\text{kg})\dfrac{(172~\text{m/s}) }{(13~\text{s})}.##
## F_{\text{th.}}=(245,000~\text{N})+(25,000~\text{kg})\dfrac{(172~\text{m/s}) }{(13~\text{s})}.##
Assuming up as negative and down as positive
##\mathbf F_{\text{th.}}+\mathbf F_{\text{g}}=M\dfrac{\Delta \mathbf{v}}{\Delta t}.##
## F_{\text{th.}}~(-\mathbf{\hat{y}})+(245,000~\text{N})~(\mathbf{\hat{y}})=(25,000~\text{kg})\dfrac{0-(172~\text{m/s})(\mathbf{\hat{y}}) }{(13~\text{s})}.##
## -F_{\text{th.}}+(245,000~\text{N})=-(25,000~\text{kg})\dfrac{(172~\text{m/s}) }{(13~\text{s})}.##
## F_{\text{th.}}=(245,000~\text{N})+(25,000~\text{kg})\dfrac{(172~\text{m/s}) }{(13~\text{s})}.##
@haruspex and I have been over this with 1d vectors several times in the past and (I think) we have agreed to disagree. When all vector directions are known, as is the case here, I favor expressing vectors in polar form, i.e. separate the magnitude, which is always positive, from the direction which can be positive or negative. Splitting the two is easier to explain to novices who are taught that vectors have magnitude (always positive) and direction. The danger here is that some students use false logic: magnitudes are always positive, a vector is a scalar times a direction, therefore a scalar is always positive.
If both the direction and magnitude of a vector are not known, then one can assume a direction and do the algebra to find the magnitude. If the magnitude turns out to be negative, then one must conclude that the direction is in the opposite direction of the one assumed. The idea is no different from what is taught as standard practice when using Kirchhoff's laws to analyze circuits: "Assume a direction for the current and if the current turns out negative, this means that it flows in the opposite direction."